Answer:
Step-by-step explanation:
The equation of a straight line can be represented in the slope-intercept form, y = mx + c
Where c = intercept
m = slope
The equation of the given line is
y = - 23x+5
Comparing it with the slope intercept equation, slope, m = -23
If a line is perpendicular to another line, the slope of the line is the negative reciprocal of the given line. This means that the slope of the line passing through the point (8, 1) is 1/23
We would determine the intercept, c by substituting m = 1/23, x = 8 and y = 1 into y = mx + c. It becomes
1 = 1/23 × 8 + c
1 = 8/23 + c
c = 1 - 8/23 = 15/23
The equation becomes
y = x/23 + 15/23
The price to earnings ratio (P/E) is an important tool in financial work. A random sample of 14 large U.S. banks (J. P. Morgan, Bank of America, and others) gave the following P/E ratios†.24 16 22 14 12 13 17 22 15 19 23 13 11 18
The sample mean is x ≈ 17.1. Generally speaking, a low P/E ratio indicates a "value" or bargain stock.
Suppose a recent copy of a magazine indicated that the P/E ratio of a certain stock index is μ = 18.
Let x be a random variable representing the P/E ratio of all large U.S. bank stocks.
We assume that x has a normal distribution and σ = 5.1.
Do these data indicate that the P/E ratio of all U.S. bank stocks is less than 18? Use α = 0.01.(a) What is the level of significance?(b) What is the value of the sample test statistic? (Round your answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Answer:
a) [tex]\alpha=0.01[/tex] is the significance level given
b) [tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]
c) Since is a one side left tailed test the p value would be:
[tex]p_v =P(Z<-0.6603)=0.2545[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=17.1[/tex] represent the mean P/E ratio for the sample
[tex]\sigma=5.1[/tex] represent the sample standard deviation for the population
[tex]n=14[/tex] sample size
[tex]\mu_o =18[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean for the P/E ratio is less than 18, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 18[/tex]
Alternative hypothesis:[tex]\mu < 18[/tex]
If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
(a) What is the level of significance?
[tex]\alpha=0.01[/tex] is the significance level given
(b) What is the value of the sample test statistic?
We can replace in formula (1) the info given like this:
[tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]
(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Since is a one side left tailed test the p value would be:
[tex]p_v =P(Z<-0.6603)=0.2545[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean for the P/E ratio is not significantly less than 18.
Solve the following equation by taking the square root 12 - 6n2 = -420 i need help
Answer:
[tex]n=\pm 6\sqrt{2}[/tex]
Step-by-step explanation:
It may work well to divide by 6, subtract 2, and multiply by -1 before you take the square root.
[tex]12-6n^2=-420\\2-n^2=-70 \qquad\text{divide by 6}\\-n^2=-72 \qquad\text{subtract 2}\\n^2=72 \qquad\text{multiply by -1}\\\\n=\pm\sqrt{36\cdot 2} \qquad\text{take the square root}\\\\n=\pm 6\sqrt{2} \qquad\text{simplify}[/tex]
Consider the number of loudspeaker announcements per day at school. Suppse thee snce of chance ofhaving 0 announcements, a 30% chance ofhaving i announcement, a 25% having 2 announcements, a 20% chance of having 3 announcements, and a \0 % chance announcements. Find the expected value of the number of announcements per day. of having A
Answer:
The expected value is 1.8
Step-by-step explanation:
Consider the provided information.
Suppose there’s a 15% chance of having 0 announcements, a 30% chance of having 1 announcement, a 25% chance of having 2 announcements, a 20% chance of having 3 announcements, and a 10% chance of having 4 announcements.
[tex]\text{Expected Value}=a \cdot P(a) + b \cdot P(b) + c \cdot P(c) + \cdot\cdot[/tex]
Where a is the announcements and P(a) is the probability.
[tex]\text{Expected Value}=0\cdot 15\% + 1 \cdot 30\% + 2 \cdot 25\% + 3\cdot20\%+4\cdot10[/tex]
[tex]\text{Expected Value}=1 \cdot 0.30+2 \cdot 0.25 +3 \cdot 0.2 + 4\cdot 0.10[/tex]
[tex]\text{Expected Value}=1.8[/tex]
Hence, the expected value is 1.8
A bank with branches located in a commercial district of a city and in a residential district has the business
objective of developing an improved process for serving customers during the noon-to-1 P.M. lunch
period. Management decides to first study the waiting time in the current process. The waiting time is
defined as the time that elapses from when the customer enters the line until he or she reaches the teller
window. Data are collected from a random sample of 15 customers at each branch.
The following is the data sample of the wait times, in minutes, from the commercial district branch.
4.14 5.66 3.04 5.34 4.82 2.69 3.32 3.41
4.42 6.01 0.15 5.11 6.59 6.43 3.72
The following is the data sample of the wait times, in minutes, from the residential district branch.
9.99 5.89 8.06 5.91 8.64 3.77 8.21 8.52
10.46 6.87 5.53 4.23 6.25 9.88 5.59
Determine the test statistic.
Answer:
test statistic is 4.27
Step-by-step explanation:
[tex]H_{0}[/tex] : mean waiting time in a residential district branch is the same as a commercial district branch
[tex]H_{a}[/tex] : mean waiting time in a residential district branch is more than a commercial district branch
commercial district branch:
mean waiting time: [tex]\frac{4.14+5.66+3.04+5.34+4.82+2.69+3.32+3.41+4.42+6.01+0.15+5.11+6.59+6.43+3.72}{15} =4.32[/tex]
standard deviation:
mean squared differences from the mean = 1.63
residential district branch.
mean waiting time: [tex]\frac{9.99+5.89+8.06+5.91+8.64+3.77+8.21+8.52+10.46+6.87+5.53+4.23+6.25+9.88+5.59}{15} =7.19[/tex]
standard deviation:
mean squared differences from the mean = 2.03
test statstic can be calculated using the formula:
[tex]z=\frac{X-Y}{\sqrt{\frac{s(x)^2}{N(x)}+\frac{s(y)^2}{N(y)}}}[/tex] where
X is the mean mean waiting time for residential district branch. (7.19)Y is the mean mean waiting time for commercial district branch. (4.32)s(x) is the sample standard deviation for residential district branch (2.03)s(y) is the sample standard deviation for commercial district branch.(1.93)N(x) is the sample size for residential district branch (15)N(y) is the sample size for commercial district branch.(15)[tex]z=\frac{7.19-4.32}{\sqrt{\frac{2.03^2}{15}+\frac{1.63^2}{15}}}[/tex] ≈4.27
A researcher matched 30 participants on intelligence (hence 15 pairs of participants), and then compared differences in emotional responsiveness to two experimental stimuli between each pair. For this test, what are the critical values, assuming a two-tailed test at a 0.05 level of significance?
(A) ±2.042
(B) ±2.045
(C) ±2.131
(D) ±2.145
We wish to obtain a 90% confidence interval for the standard deviation of a normally distributed random variable. To accomplish this we obtain a simple random sample of 16 elements from the population on which the random variable is defined. We obtain a sample mean value of 20 with a sample standard deviation of 12. Give the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable. a) 83 to 307 b) 9 to 18 c) 91 to 270 d) 15 to 25 e) 20 to 34
Answer: d) 15 to 25
Step-by-step explanation:
Given : Sample size : n= 16
Degree of freedom = df =n-1 = 15
Sample mean : [tex]\overline{x}=20[/tex]
sample standard deviation : [tex]s= 12[/tex]
Significance level : [tex]\alpha= 1-0.90=0.10[/tex]
Since population standard deviation is unavailable , so the confidence interval for the population mean is given by:-
[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
Using t-distribution table , we have
Critical value = [tex]t_{\alpha/2, df}=t_{0.05 , 15}=1.7530[/tex]
90% confidence interval for the mean value will be :
[tex]20\pm (1.7530)\dfrac{12}{\sqrt{16}}[/tex]
[tex]20\pm (1.7530)\dfrac{12}{4}[/tex]
[tex]20\pm (1.7530)(3)[/tex]
[tex]20\pm (5.259)[/tex]
[tex](20-5.259,\ 20+5.259)[/tex]
[tex](14.741,\ 25.259)\approx(15,\ 25 )[/tex][Round to the nearest integer]
Hence, the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable.= 15 to 25.
Final answer:
To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable with a sample size of 16, sample mean of 20, and sample standard deviation of 12, use the chi-square distribution to calculate the lower and upper bounds. The 90% confidence interval is approximately 86 to 283.
Explanation:
To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable, we can use the chi-square distribution. Given a simple random sample of 16 elements with a sample mean of 20 and a sample standard deviation of 12, we can calculate the lower and upper bounds of the confidence interval.
Step 1: Calculate the chi-square values for the lower and upper bounds using the following formulas:
Lower bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 90% confidence level with (n-1) degrees of freedom.
Upper bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 10% significance level with (n-1) degrees of freedom.
Substituting the values into the formulas, we get:
Lower bound: (15)(144) / 24.996 = 86.437
Upper bound: (15)(144) / 7.633 = 283.368
Rounding to the nearest integer, the 90% confidence interval for the standard deviation of the random variable is approximately 86 to 283.
A marketing company is interested in the proportion of people that will buy a particular product. Match the vocabulary word with its corresponding example. The 380 randomly selected people who are observed to see if they will buy the product The proportion of the 380 observed people who buy the product fAll people in the marketing company's region The list of the 380 Yes or No answers to whether the person bought the product The proportion of all people in the company's region who buy the product Purchase: Yes or No whether a person bought the product a. Statistic b. Data Sample d. Variable e. Parameter f. Population Points possible: 6 License
The matching is as follow:
a -> Statistic
b -> Data Sample
d -> Variable
e -> Parameter
f -> Population
a. Statistic: The proportion of the 380 observed people who buy the product
b. Data Sample: The 380 randomly selected people who are observed to see if they will buy the product
d. Variable: Purchase - Yes or No whether a person bought the product
e. Parameter: The proportion of all people in the company's region who buy the product
f. Population: All people in the marketing company's region
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The 380 randomly selected people are the 'Data Sample', the proportion of these who buy is a 'Statistic', all people in the region are the 'Population', the list of 380 Yes/No answers is the 'Variable', proportion of all people in the region who buy the product is 'Parameter', and yes/no answer for each person's purchase is also deemed a 'Variable'.
Explanation:In this question, we are dealing with terms related to statistic studies. The 380 randomly selected people who are observed to see if they will buy the product represent the Data Sample. The proportion of the 380 observed people who buy the product is considered a Statistic. All people in the marketing company's region is the Population. The list of the 380 Yes or No answers to whether the person bought the product constitutes the Variable. The proportion of all people in the company's region who buy the product is an example of a Parameter. Lastly, the Purchase: Yes or No whether a person bought the product is the Variable.
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Before lending someone money, banks must decide whether they believe the applicant will repay the loan. One strategy used is a point system. Loan officers assess information about the applicant, totalling points they award for the persons income level, credit history, current debt burden, and so on. The higher the point total, the more convinced the bank is that it’s safe to make the loan. Any applicant with a lower point total than a certain cut-off score is denied a loan. We can think of this decision as a hypothesis test. Since the bank makes its profit from the interest collected on repaid loans, their null hypothesis is that the applicant will repay the loan and therefore should get the money. Only if the persons score falls below the minimum cut-off will the bank reject the null and deny the loan. This system is reasonably reliable, but, of course, sometimes there are mistakes.a) When a person defaults on a loan, which type of error did the bank make?b) Which kind of error is it when the bank misses an opportunity to make a loan to someone who would have repaid it?c) Suppose the bank decides to lower the cut-off score from 250 points to 200. Is that analogous to choosing a higher or lower value of for a hypothesis test? Explain.d) What impact does this change in the cut-off value have on the chance of each type of error?
Answer:
(a) Type II error
(b) Type I error
(c) It is analogous to choosing a lower value for a hypothesis test
(d) There will be more tendency of making type II error and less tendency of making type I error
Step-by-step explanation:
(a) The bank made a type II error because they accepted the null hypothesis when it is false
(b) The bank made a type I error because they rejected the null hypothesis when it is true
(c) By lowering the value for the hypothesis test, they give applicants who do not meet the initial cut-off point the benefit of doubt of repaying the loan thus increasing their chances of making more profit
(d) There will be more tendency of making type II error because the bank accepts the null hypothesis though they are not fully convinced the applicants will repay the loan and less tendency of making type I error because the bank rejects the null hypothesis knowing the applicants might not be able to repay the loan
In hypothesis testing, a person defaulting on a loan represents a Type I error, while missing an opportunity to make a loan to someone who would have repaid it represents a Type II error. Lowering the cut-off score is analogous to increasing the value in a hypothesis test, accepting more risk. This increases the likelihood of Type I errors but decreases the likelihood of Type II errors.
Explanation:In the context of hypothesis testing in banking and the financial capital market, (a) when a person defaults on a loan, the bank made a Type I error: they lent money to an individual who failed to repay it. (b) If the bank does not lend money to someone who would have repaid it, it's a Type II error: they missed an opportunity to profit from interest because they incorrectly predicted the person would not pay back the loan. (c) Lowering the cut-off score from 250 points to 200 is analogous to choosing a higher value for a hypothesis test, which means the bank is willing to accept more risk. (d) Changing the cut-off value impacts the chance of each kind of error. By lowering the score, the bank is more likely to make Type I errors (lending to individuals who won't repay), but less likely to make Type II errors (not lending to individuals who would repay).
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In the following sequence, each number (except the first two) is the sum of the previous two numbers: 0, 1, 1, 2, 3, 5, 8, 13, .... This sequence is known as the Fibonacci sequence. We speak of the i'th element of the sequence (starting at 0)-- thus the 0th element is 0, the 1st element is 1, the 2nd element is 1, the 3rd element is 2 and so on. Given the positive integer n, associate the nth value of the fibonacci sequence with the variable result. For example, if n is associated with the value 8 then result would be associated with 21.
Final answer:
To find the nth Fibonacci number, dynamic programming stores previously calculated values in an array, which allows for efficient linear time computation by summing the two previous numbers to obtain the nth value.
Explanation:
The Fibonacci sequence is defined such that each number in the sequence is the sum of the two preceding ones, starting from 0 and 1. To calculate the nth Fibonacci number, denoted as Fib(n), we start by setting Fib(0) and Fib(1) equal to 0 and 1, respectively. For n ≥ 2, Fib(n) is defined recursively as Fib(n) = Fib(n - 1) + Fib(n - 2). A naive recursive algorithm could be inefficient due to repeated calculations. Using dynamic programming or memoization improves efficiency by storing intermediate results, thus avoiding unnecessary recalculations.
Computing Fibonacci Numbers Using Dynamic Programming
To compute the nth Fibonacci number using dynamic programming, we create an array or list to save previously computed Fibonacci numbers. The nth value, for instance Fib(8) = 21, is then easily found by summing up the n-1th and n-2th values from the array, which are already computed and stored. This approach leads to a time complexity that is linear, i.e., O(n), instead of exponential.
The weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds. What's the probability that a person weighs more than 200 pounds? (round your answer to the nearest hundredth)
Answer:
0.23.
Step-by-step explanation:
We have been given that the weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds.
First of all, we will find the z-score corresponding to sample score 200 using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=[/tex] Z-score,
[tex]x=[/tex] Sample score,
[tex]\mu=[/tex] Mean,
[tex]\sigma=[/tex] Standard deviation.
[tex]z=\frac{200-185}{20}[/tex]
[tex]z=\frac{15}{20}[/tex]
[tex]z=0.75[/tex]
Now, we need to find [tex]P(z>0.75)[/tex]. Using formula [tex]P(z>a)=1-P(z<a)[/tex], we will get:
[tex]P(z>0.75)=1-P(z<0.75)[/tex]
Using normal distribution table, we will get:
[tex]P(z>0.75)=1-0.77337 [/tex]
[tex]P(z>0.75)=0.22663 [/tex]
Round to nearest hundredth:
[tex]P(z>0.75)\approx 0.23[/tex]
Therefore, the probability that a person weighs more than 200 pounds is approximately 0.23.
Answer:the probability that a person weighs more than 200 pounds is 0.23
Step-by-step explanation:
Since the weight of people on a college campus are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = weight of people on a college campus
u = mean weight
s = standard deviation
From the information given,
u = 185
s = 20
We want to find the probability that a person weighs more than 200 pounds. It is expressed as
P(x greater than 200) = P(x greater than 200) = 1 - P(x lesser than lesser than or equal to 200).
For x = 200,
z = (200 - 185)/20 = 0.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.7735
P(x greater than 200) = 1 - 0.7735 = 0.23
Identify the type of observational study (cross-sectional, retrospective, or prospective) described below. A research company uses a device to record the viewing habits of about 2500 households, and the data collected over the past 2 years will be used to determine whether the proportion of households tuned to a particular children's program increased. Which type of observational study is described in the problem statement?
A. A prospective study
B. A retrospective study
C. A cross-sectional study
Answer:
B
Step-by-step explanation:
The retrospective or historic cohort story, is a longitudinal cohort story that considers a particular set of individuals that share the same exposure factor to ascertain its influence in the developments of an occurrence, which are compared with the other set or cohort which were not exposed to the same factors.
Retrospective studies have existed about the same time as prospective studies, hence their names.
In order to determine whether or not there is a significant difference between the hourly wages of two companies, the following data have been accumulated.
Company 1 Company 2 n1 = 80 n2 = 60 x̄1 = $10.80 x̄2 = $10.00 σ1 = $2.00 σ2 = $1.50 Refer to Exhibit 10-13. The point estimate of the difference between the means (Company 1 – Company 2) is _____.
a. .8
b. –20
c. .50
d. 20
Answer:
a. .8
Step-by-step explanation:
The point estimate of the difference between the means of Company 1 and Company 2 can be calculated as:
point estimate = mu1 - mu2 where
mu1 is the sample mean hourly wage of Company 1mu2 is the sample mean hourly wage of Company 2Therefore point estimate = $10.80- $10 =$ .8
One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Construct a 99% confidence interval for the population mean hours spent watching television per month.
Fill in the blank: Round to two decimal places. ( , )
Answer: (143.07, 158.93)
Step-by-step explanation:
The formula to find the confidence interval is given by :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
where n= sample size
[tex]\overline{x}[/tex] = Sample mean
z* = critical z-value (two tailed).
[tex]\sigma[/tex] = Population standard deviation
We assume that the underlying population distribution is normal.
As per given , we have
n= 108
[tex]\overline{x}=151[/tex]
[tex]\sigma=32[/tex]
Critical value for 99% confidence level = 2.576 (By using z-table)
Then , the 99% confidence interval for the population mean hours spent watching television per month :-
[tex]151\pm (2.576)\dfrac{32}{\sqrt{108}}[/tex]
[tex]151\pm (2.576)\dfrac{32}{10.3923048454}[/tex]
[tex]151\pm (2.576)(3.07920143568)[/tex]
[tex]151\pm (7.93202289831)\approx151\pm7.93\\\\=(151-7.93,\ 151+7.93)\\\\=(143.07,\ 158.93 )[/tex]
Hence, the required 99% confidence interval for the population mean hours spent watching television per month. = (143.07, 158.93)
The 99% confidence interval for the average number of hours all Americans spend watching television per month, based on the given sample, is (143.76, 158.23). This is computed using the confidence interval formula with the given sample mean, standard deviation, and the z-score for a 99% confidence interval.
Explanation:The question involves the concept of the confidence interval in statistics. Here we are given the sample size (n=108), the sample mean ([tex]\overline{X}[/tex] = 151), and the sample standard deviation (s=32). We are required to compute the 99% confidence interval.
To calculate a confidence interval, we apply this formula: [tex]\overline{X}[/tex] ± (z-value * (s/√n)) Where '[tex]\overline{X}[/tex]' is the sample mean, 'z-value' is the Z-score (which for a 99% confidence interval is 2.58), 's' is the standard deviation and 'n' is the sample size.
Substitute the given values into the formula: 151 ± (2.58 * (32/√108))
This results in: (143.76, 158.23)
So, we can say with 99% confidence that the average number of hours all Americans spend watching television per month is between 143.76 hours and 158.23 hours.
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Identify the sampling technique used. In a recent television survey, participants were asked to answer "yes" or "no" to the question "Are you in favor of the death penalty?" Six thousand five hundred responded "yes" while 51 00 responded "no". There was a fifty- cent charge for the call.
Answer:
Convenience sampling. See explanation below.
Step-by-step explanation:
For this case they not use random sampling since all the individuals for the population are not included on the sampling frame, some individuals have probability of inclusion 0, because they are just using a charge for the call and some people would not answer the call.
Is not stratified sampling since we don't have strata clearly defined on this case, and other important thing is that in order to apply this method we need homogeneous strata groups and that's not satisfied on this case.
Is not systematic sampling since they not use a random number or a random starting point, and is not mentioned, they just use a call that is charge with 50 cents.
Is not cluster sampling since we don't have clusters clearly defined, and again in order to apply this method we need homogeneous characteristics on the clusters and an equal chance of being a part of the sample, and that's not satisfid again with the call charge used.
So then the only method that can be possible for this case is convenience sampling because they use a non probability sampling with some members of the potential population with probability of inclusion 0.
The sampling technique used in the given scenario is voluntary response sampling, where participants decide whether to take part in the survey. In this technique, participants chose to respond to the television survey by making a call. This method can be biased as the responses could lean towards those who hold strong views on the topic.
Explanation:The sampling technique used in this scenario is referred as voluntary response sampling or self-selection sampling. In this method, participants themselves decide to participate or not, usually by responding to a call for participants. This often happens when surveys are disseminated widely such as through television or online. Since there was a call to answer "yes" or "no" for the question with a charge, individuals chose to participate by making a call. It is important to note that the main drawback of this technique is that it tends to be biased, as the sample could be skewed in favor of those who felt strongly about the topic.
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A linear enzyme is formed by four alpha and two beta protein subunits. How manydifferent arrangements are there?
Answer:
15
Step-by-step explanation:
We are given that
Number of alpha protein subunits=4
Number of beta protein subunits=2
Total number of protein sub-units=2+4=6
We have to find the number of different arrangements are there.
When r identical letters and y identical letters and total object are n then arrangements are
[tex]\frac{n!}{r!x!}[/tex]
n=6,r=2,x=4
By using the formula
Then, we get
Number of different arrangements =[tex]\frac{6!}{2!4!}[/tex]
Number of different arrangements=[tex]\frac{6\times 5\times 4!}{2\times 1\times 4!}[/tex]
Number of different arrangements=15
Hence, different arrangements are there= 15
A particle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, where t is measured in seconds and s in feet.
(a) Find the velocity at time t.
(b) What is the velocity after 3 seconds?
(c) When is the particle at rest?
Answer:
a) [tex]v(t) = 3t^{2} - 16t + 2[/tex]
b) The velocity after 3 seconds is -3m/s.
c) [tex]t = 0.13s[/tex] and [tex]t = 5.2s[/tex].
Step-by-step explanation:
The position is given by the following equation.
[tex]s(t) = t^{3} - 8t^{2} + 2t[/tex]
(a) Find the velocity at time t.
The velocity is the derivative of position. So:
[tex]v(t) = s^{\prime}(t) = 3t^{2} - 16t + 2[/tex].
(b) What is the velocity after 3 seconds?
This is v(3).
[tex]v(t) = 3t^{2} - 16t + 2[/tex]
[tex]v(3) = 3*(3)^{2} - 16*(3) + 2 = -19[/tex]
The velocity after 3 seconds is -3m/s.
(c) When is the particle at rest?
This is when [tex]v(t) = 0[/tex].
So:
[tex]v(t) = 3t^{2} - 16t + 2[/tex]
[tex]3t^{2} - 16t + 2 = 0[/tex]
This is when [tex]t = 0.13s[/tex] and [tex]t = 5.2s[/tex].
An article reported that for a sample of 58 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 165.4.
(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) , ppm Interpret the resulting interval. We are 95% confident that the true population mean lies below this interval. We are 95% confident that this interval does not contain the true population mean. We are 95% confident that this interval contains the true population mean. We are 95% confident that the true population mean lies above this interval.
(b) Suppose the investigators had made a rough guess of 184 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 47 ppm for a confidence level of 95%?
Answer:
Step-by-step explanation:
Twenty years ago, entering male high school students of Central High could do an average of 24 pushups in 60 seconds. To see whether this remains true today, a random sample of 36 freshmen was chosen. Suppose their average was 22.5 with a sample standard deviation of 3.1,
(a) Test, using the p-value approach, whether the mean is still equal to 24 at the 5 percent level of significance.
(b) Calculate the power of the test if the true mean is 23.
Answer:
a) [tex]t=\frac{22.5-24}{\frac{3.1}{\sqrt{36}}}=-2.903[/tex]
[tex]p_v =2*P(t_{(35)}<-2.903)=0.0064[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.
b) Power =0.4626+0.000172=0.463
See explanation below.
Step-by-step explanation:
Part a
Data given and notation
[tex]\bar X=22.5[/tex] represent the sample mean
[tex]s=3.1[/tex] represent the sample standard deviation
[tex]n=36[/tex] sample size
[tex]\mu_o =24[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is still equal to 24, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 24[/tex]
Alternative hypothesis:[tex]\mu \neq 24[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{22.5-24}{\frac{3.1}{\sqrt{36}}}=-2.903[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=36-1=35[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{(35)}<-2.903)=0.0064[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.
Part b
The Power of a test is the probability of rejecting the null hypothesis when, in the reality, it is false.
For this case the power of the test would be:
P(reject null hypothesis| [tex]\mu=23[/tex])
If we see the null hypothesis we reject it when we have this:
The critical values from the t distribution with 35 degrees of freedom and at 5% of significance are -2.03 and 2.03. From the z score formula:
[tex]t=\frac{\bar x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
If we solve for [tex]\bar x[/tex] we got:
[tex]\bar X= \mu \pm t \frac{s}{\sqrt{n}}[/tex]
Using the two critical values we have the critical values four our sampling distribution under the null hypothesis
[tex]\bar X= 24 -2.03 \frac{3.1}{\sqrt{36}}=22.951[/tex]
[tex]\bar X= 24 +2.03 \frac{3.1}{\sqrt{36}}=25.049[/tex]
So we reject the null hypothesis if [tex]\bar x<22.951[/tex] or [tex]\bar X >25.049[/tex]
So for our case:
P(reject null hypothesis| [tex]\mu=23[/tex]) can be founded like this:
[tex]P(\bar X <22.951|\mu=23)=P(t<\frac{22.951-23}{\frac{3.1}{\sqrt{36}}})=P(t_{35}<-0.0948)=0.4626[/tex]
[tex]P(\bar X >25.012|\mu=23)=P(t<\frac{25.049-23}{\frac{3.1}{\sqrt{36}}})=P(t_{35}>3.966)=0.000172[/tex]
And the power on this case would be the sum of the two last probabilities:
Power =0.4626+0.000172=0.463
Final answer:
To test whether the mean length of time students spend doing homework each week has increased, we can conduct a hypothesis test using the null hypothesis that the mean time is still 2.5 hours and the alternative hypothesis that the mean time has increased.
Explanation:
To test whether the mean length of time students spend doing homework each week has increased, we can conduct a hypothesis test. The null hypothesis, denoted as H0, would be that the mean time is still 2.5 hours. The alternative hypothesis, denoted as Ha, would be that the mean time has increased. In this case, the alternative hypothesis would be Ha: µ > 2.5, where µ represents the population mean.
To conduct the hypothesis test, we can use a t-distribution because the population standard deviation is not known. We can calculate the test statistic by using the formula: t = (x - µ) / (s/√n), where x is the sample mean, µ is the hypothesized mean, s is the sample standard deviation, and n is the sample size. Once we calculate the test statistic, we can compare it to the critical value from the t-distribution table or calculate the p-value to determine the level of significance.
Suppose that your company has just developed a new screening test for a disease and you are in charge of testing its validity and feasibility.You decide to evaluate the test on 1000 individuals and compare the results of the new test to the gold standard.You know the prevalence of disease in your population is 30%.The screening test gave a positive result for 292 individuals.285 of these individuals actually had the disease on the basis of the gold standard determination.
Calculate the sensitivity of the new screening test.
95.0% 97.6% 99.0% 96.9%
Answer:
The sensitivity of the new screening test is 97.6%
Step-by-step explanation:
The sensitivy of a test, or true positive rate, is defined as the proportion of positive results that are correctly identified. It is complementary to the proportion of "false positives".
In this case the test gave 292 positive results. Of this 292 tests, 285 of these individuals actually had the disease.
So the sensititivity is equal to the ratio of the true positives (285) and the total positives (292)
[tex]Sensitivity=\frac{TP}{P}=\frac{285}{292}=0.976=97.6\%[/tex]
Over the past semester, you've collected the following data on the time it takes you to get to school by bus and by car:
• Bus:(15,10,7,13,14,9,8,12,15,10,13,13,8,10,12,11,14,11,9,12) • Car:(5,8,7,6,9,12,11,10,9,6,8,10,13,12,9,11,10,7)
You want to know if there's a difference in the time it takes you to get to school by bus and by car.
A. What test would you use to look for a difference in the two data sets, and what are the conditions for this test? Do the data meet these conditions? Use sketches of modified box-and-whisker plots to support your decision.
B. What are the degrees of freedom (k) for this test using the conservative method? (Hint: Don't pool and don't use your calculator.)
C. What are the sample statistics for this test? Consider the data you collected for bus times to be sample one and the data for car times to be sample two.D. Compute a 99% confidence interval for the difference between the time it takes you to get to school on the bus and the time it takes to go by car. Draw a conclusion about this difference based on this confidence interval using
E. Constructthesameconfidenceintervalyoudidinpartd,thistimeusingyour graphing calculator. Show what you do on your calculator, and what you put into your calculator, and give the confidence interval and degrees of freedom. (Hint: Go back to previous study materials for this unit if you need to review how to do this.)
F. How is the interval computed on a calculator different from the interval computed by hand? Why is it different? In this case, would you come to a different conclusion for the hypothesis confidence interval generated by the calculator?
Answer:
Step-by-step explanation:
Hello!
You have two study variables
X₁: Time it takes to get to school by bus.
X₂: Time it takes to get to school by car.
Data:
Sample 1
Bus:(15,10,7,13,14,9,8,12,15,10,13,13,8,10,12,11,14,11,9,12)
n₁= 20
Mean X[bar]₁= 11.30
S₁= 2.39
Sample 2
Car:(5,8,7,6,9,12,11,10,9,6,8,10,13,12,9,11,10,7)
n₂= 18
Mean X[bar]₂= 9.06
S₂= 2.29
A.
To test if there is any difference between the times it takes to get to school using the bus or a car you need to compare the means of each population.
The condition needed to make a test for the difference between means is that both the independent population should have a normal distribution.
The sample sizes are too small to use an approximation with the CLT. You can test if the study variables have a normal distribution using different methods, and hypothesis test, using a QQ-plot or using the Box and Whiskers plot. The graphics are attached.
As you can see both samples show symmetric distribution, the boxes are proportioned, the second quantile (median) and the mean (black square) are similar and in the center of the boxes. The whiskers have the same length and there are no outliers. Both plots show symmetry centered in the mean consistent with a normal distribution. According to the plots you can assume both variables have a normal distribution.
The next step to select the statistic to test the population means is to check whether there is other population information available.
If the population variances are known, you can use the standard normal distribution.
If the population variances are unknown, the distribution to use is a Student's test.
If the unknown population variances are equal, you can use a t-test with a pooled sample variance.
If the unknown population variances are not equal, the t-test to use is the Welch approximation.
Using an F-test for variance homogeneity the p-value is 0.43 so at a 0.01 level, you can conclude that the population variances are equal.
The statistic to use is a pooled t-test.
B.
Degrees of freedom.
For each study variable, you can use a t-test with n-1 degrees of freedom.
For X₁ ⇒ n₁-1 = 20 - 1 = 19
For X₂ ⇒ n₂-1 = 18 = 17
For X₁ + X₂ ⇒ (n₁-1) + (n₂-1)= n₁ + n₂ - 2= 20 + 18 - 2= 36
C.
See above.
D.
The formula for the 99% confidence interval is:
(X[bar]₁ - X[bar]₂) ± [tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex] * [tex]Sa\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }[/tex]
[tex]Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} }[/tex]
[tex]Sa= \sqrt{\frac{19*(2.39)^2+17*(2.29)^2}{36} }[/tex]
Sa= 2.34
[tex]t_{n_1+n_2-2; 1- \alpha /2}[/tex]
[tex]t_{36; 0.995}[/tex] = 2.72
(11.30 - 9.06) ± 2.72 * [tex]2.34\sqrt{\frac{1}{20} + \frac{1}{18} }[/tex]
[0.17;4.31]
With a 99% confidence level you'd expect that the difference between the population means of the time that takes to get to school by bus and car is contained in the interval [0.17;4.31].
E.
Couldn't find the original lesson to see what calculator is used.
F.
Same, no calculator available.
I hope it helps!
Answer:
this is nnot the answer i was looking for
Step-by-step explanation:
A least squares regression line was found. Using technology, it was determined that the total sum of squares (SST) was 46.8 and the sum of squares of regression (SSR) was 14.55. Use these values to calculate the percent of the variability in y that can be explained by variability in the regression model. Round your answer to the nearest integer.
Answer: 31%
Step-by-step explanation:
Formula : Percent of the variability = [tex]R^2\times100=\dfrac{SSR}{SST}\times100[/tex]
, where [tex]R^2[/tex] = Coefficient of Determination.
SSR = sum of squares of regression
SST = total sum of squares
[tex]R^2[/tex] is the proportion of the variation of Y that can be attributed to the variation of x.
As per given , we have
SSR = 14.55
SST= 46.8
Then, the percent of the variability in y that can be explained by variability in the regression model =[tex]\dfrac{14.55}{46.8}\times100=31.0897435897\%\approx31\%[/tex]
Hence, the percent of the variability in y that can be explained by variability in the regression model = 31%
Answer: 14.55/46.8= .3109
.3109x100=31.09
Step-by-step explanation:
We know that narrower confidence intervals give us a more precise estimate of the true population proportion. Which of the following could we do to produce higher precision in our estimates of the population proportion?
A. We can select a lower confidence level and increase the sample size.
B. We can select a higher confidence level and decrease the sample size.
C. We can select a higher confidence level and increase the sample size.
D. We can select a lower confidence level and decrease the sample size.
Answer:
A. We can select a lower confidence level and increase the sample size.
Step-by-step explanation:
The length of a confidence interval is:
Direct proportional to the confidence interval. This means that the higher the confidence level, the higher the length of the interval is.
Inverse proportional to the size of the sample.This means that the higher the size of the sample, the lower, or narrower, the length of the interval is.
Which of the following could we do to produce higher precision in our estimates of the population proportion?
We want a narrower interval. So the correct answer is:
A. We can select a lower confidence level and increase the sample size.
Automated manufacturing operations are quite precise but still vary, often with distributions that are close to Normal. The width in inches of slots cut by a milling machine follows approximately the N(0.8750, 0.0012) distribution. The specifications allowslot widths between 0.8725 and 0.8775 inch. What proportion of slots meet these specifications?
Answer:
96.2% of slots meet these specifications.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.8750
Standard Deviation, σ = 0.0012
We are given that the distribution of width in inches of slots is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P( widths between 0.8725 and 0.8775 inch)
[tex]P(0.8725 \leq x \leq 0.8775) = P(\displaystyle\frac{0.8725 - 0.8750}{0.0012} \leq z \leq \displaystyle\frac{0.8775-0.8750}{0.0012}) = P(-2.083 \leq z \leq 2.083)\\\\= P(z \leq 2.083) - P(z < -2.083)\\= 0.981 - 0.019 = 0.962 = 96.2\%[/tex]
[tex]P(0.8725 \leq x \leq 0.8775) = 96.2\%[/tex]
96.2% of slots meet these specifications.
The question asks for the proportion of slots meeting width specifications within a normal distribution with defined mean and standard deviation. We calculate the corresponding Z-scores for the lower and upper specification limits and then determine the probability of a slot falling within these limits.
Explanation:The problem involves finding the proportion of slots that meet the specified width requirements in a normal distribution. In this case, the slot widths follow a normal distribution with a mean (μ) of 0.8750 inches, and a standard deviation (σ) of 0.0012 inches. The specifications require that slot widths be between 0.8725 inches and 0.8775 inches.
To find the proportion of slots that meet these specifications, we calculate the Z-scores for both the lower specification limit of 0.8725 and the upper specification limit of 0.8775. The Z-score formula is given by Z = (X - μ) / σ, where X is the value for which we want to find the Z-score.
For the lower limit, we have:
Z(lower) = (0.8725 - 0.8750) / 0.0012 = -2.083…
For the upper limit, we have:
Z(upper) = (0.8775 - 0.8750) / 0.0012 = 2.083…
Next, we use the standard normal distribution to find the probability corresponding to these Z-scores. The area under the curve between these two Z-scores represents the proportion of slots that are within the specifications. This can be found using standard normal distribution tables or a calculator with statistical functions.
A company wants to determine where they should locate a new warehouse. They have two existing production plants (i.e., Plant A and Plant B) that will ship units of a product to this warehouse. Plant A is located at the (X, Y) coordinates of (50, 100) and will have volume of shipping of 250 units a day. Plant B is located at the (X, Y) coordinates of (150, 200) and will have a volume of shipping of 150 units a day. Using the centroid method, which of the following are the X and Y coordinates for the new plant location?
Answer:
X = 87.5
Y = 137.5
Step-by-step explanation:
Let's X and Y be the xy-coordinates of the center warehouse.
We know that X is in between the x coordinates or the 2 plants:
50 < X < 150
Similarly Y is in between the y coordinates or the 2 plants:
100 < Y < 200
Using centroid method with the shipping units being weight we can have the following equations
250*50 + 150*150 = X*(250 + 150)
Hence X = (250*50 + 150*150)/(250+150) = 87.5
Similarly 250*100 + 150*200 = Y*(250 + 150)
Hence Y = (250*100 + 150*200)/(250+150) = 137.5
company manufactures and sells x cellphones per week. The weekly price-demand and cost equations are given below. p equals 500 minus 0.5 xp=500−0.5x and Upper C (x )equals 25 comma 000 plus 140 xC(x)=25,000+140x (A) What price should the company charge for the phones, and how many phones should be produced to maximize the weekly revenue
Answer:
The number of cellphones to be produced per week is 500.
The cost of each cell phone is $250.
The maximum revenue is $1,25,000
Step-by-step explanation:
We are given the following information in the question:
The weekly price-demand equation:
[tex]p(x)=500-0.5x[/tex]
The cost equation:
[tex]C(x) = 25000+140x[/tex]
The revenue equation can be written as:
[tex]R(x) = p(x)\times x\\= (500-0.5x)x\\= 500x - 0.5x^2[/tex]
To find the maximum value of revenue, we first differentiate the revenue function:
[tex]\displaystyle\frac{dR(x)}{dx} = \frac{d}{dx}(500x - 0.5x^2) = 500-x[/tex]
Equating the first derivative to zero,
[tex]\displaystyle\frac{dR(x)}{dx} = 0\\\\500-x = 0\\x = 500[/tex]
Again differentiating the revenue function:
[tex]\displaystyle\frac{dR^2(x)}{dx^2} = \frac{d}{dx}(500 - x) = -1[/tex]
At x = 500,
[tex]\displaystyle\frac{dR^2(x)}{dx^2} < 0[/tex]
Thus, by double derivative test, R(x) has the maximum value at x = 500.
So, the number of cellphones to be produced per week is 500, in order to maximize the revenue.
Price of phone:
[tex]p(500)=500-0.5(500) = 250[/tex]
The cost of each cell phone is $250.
Maximum Revenue =
[tex]R(500) = 500(500) - 0.5(500)^2 = 125000[/tex]
Thus, the maximum revenue is $1,25,000
A lab technician is tested for her consistency by making multiple measurements of the cholesterol level in one blood sample. The target precision is a standard deviation of 1.2 mg/dL or less. If 12 measurements are taken and the standard deviation is 1.8 mg/dL, is there enough evidence to support the claim that her standard deviation is greater than the target, at = .01? (Show the answers to all 5 steps of the hypothesis test.)
Step-by-step explanation:
Given precision is a standard deviation of s=1.8, n=12, target precision is a standard deviation of σ=1.2
The test hypothesis is
H_o:σ <=1.2
Ha:σ > 1.2
The test statistic is
chi square = [tex]\frac{(n-1)s^2}{\sigma^2}[/tex]
=[tex]\frac{(12-1)1.8^2}{1.2^2}[/tex]
=24.75
Given a=0.01, the critical value is chi square(with a=0.01, d_f=n-1=11)= 3.05 (check chi square table)
Since 24.75 > 3.05, we reject H_o.
So, we can conclude that her standard deviation is greater than the target.
Evaluate the integral Integral from nothing to nothing ∫ StartFraction 3 Over t Superscript 4 EndFraction 3 t4 sine left parenthesis StartFraction 1 Over t cubed EndFraction minus 6 right parenthesis sin 1 t3 −6dt
Answer:
[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]
Step-by-step explanation:
Given function:
[tex]\int {\frac{3}{t^4}\sin (\frac{1}{t^3}-6)} \, dt[/tex]
Now,
let [tex]\frac{1}{t^3}-6[/tex] be 'x'
Therefore,
[tex]d(\frac{1}{t^3}-6)[/tex] = dx
or
[tex]\frac{-3}{t^4}dt[/tex] = dx
on substituting the above values in the equation, we get
⇒ ∫ - sin (x) . dx
or
⇒ cos (x) + c [ ∵ ∫sin (x) . dx = - cos (x)]
Here,
c is the integral constant
on substituting the value of 'x' in the equation, we get
[tex]\cos (\frac{1}{t^3}-6)} + c[/tex]
Find the percent of the data that can be explained by the regression line and regression equation given that the correlation coefficient = -.72 (Give your answer as a percent rounded to the hundredth decimal place. Include the % sign)
Answer:
51.84%
Step-by-step explanation:
The percentage of data explained by regression line is assessed using R-square. Here, in the given scenario correlation coefficient r is given. We simply take square of correlation coefficient to get r-square. r-square=(-0.72)^2=0.5184=51.84%
The lifetime of a cheap light bulb is an exponential random variable with mean 36 hours. Suppose that 16 light bulbs are tested and their lifetimes measured. Use the central limit theorem to estimate the probability that the sum of the lifetimes is less than 600 hours.
Answer:
[tex] P(T<600)=P(Z< \frac{600-576}{144})=P(Z<0.167)=0.566[/tex]
Step-by-step explanation:
Previous concepts
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}, x>0[/tex]
And 0 for other case. Let X the random variable that represent "The number of years a radio functions" and we know that the distribution is given by:
[tex]X \sim Exp(\lambda=\frac{1}{16})[/tex]
Or equivalently:
[tex]X \sim Exp(\mu=16)[/tex]
Solution to the problem
For this case we are interested in the total T, and we can find the mean and deviation for this like this:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}=\frac{T}{n}[/tex]
If we solve for T we got:
[tex] T= n\bar X[/tex]
And the expected value is given by:
[tex] E(T) = n E(\bar X)= n \mu= 16*36=576[/tex]
And we can find the variance like this:
[tex] Var(T) = Var(n\bar X)=n^2 Var(\bar X)= n^2 *\frac{\sigma^2}{n}=n \sigma^2[/tex]
And then the deviation is given by:
[tex]Sd(T)= \sqrt{n} \sigma=\sqrt{16} *36=144[/tex]
And the distribution for the total is:
[tex] T\sim N(n\mu, \sqrt{n}\sigma)[/tex]
And we want to find this probability:
[tex] P(T< 600)[/tex]
And we can use the z score formula given by:
[tex]z=\frac{T- \mu_T}{\sigma_T}[/tex]
And replacing we got this:
[tex] P(T<600)=P(Z< \frac{600-576}{144})=P(Z<0.167)=0.566[/tex]
Using the central limit theorem, the probability that the sum of the lifetimes of 16 light bulbs is less than 600 hours is found to be approximately 0.2514 after calculating the mean, standard deviation, and z-score for the sum.
Explanation:To estimate the probability that the sum of the lifetimes of 16 light bulbs is less than 600 hours, we can use the central limit theorem. This theorem suggests that the sum (or average) of a large number of independent and identically distributed random variables will be approximately normally distributed, regardless of the original distribution of the variables. Here, each light bulb's lifetime is an exponential random variable with a mean of 36 hours.
First, we need to determine the mean (μ) and standard deviation (σ) of the sum of the lifetimes. For one light bulb, the mean is 36 hours, and since the standard deviation for an exponential distribution is equal to its mean, it is also 36 hours. For 16 light bulbs, the mean of the sum is 16 * 36 = 576 hours, and the standard deviation of the sum is √16 * 36 = 144 hours due to the square root rule for variances of independent sums.
To find the probability that the sum is less than 600 hours, we convert this to a standard normal distribution problem by calculating the z-score:
Z = (X - μ) / (σ/sqrt(n))
Z = (600 - 576) / (144/sqrt(16))
Z = 24 / 36
Z = 0.67
Now we look up the cumulative probability for a z-score of 0.67 using a standard normal distribution table or a calculator with normal distribution functions. The probability associated with a z-score of 0.67 is approximately 0.7486. Therefore, the probability that the sum of the lifetimes is less than 600 hours is 1 - 0.7486 = 0.2514.
A solid lies between planes perpendicular to the x-axis at x=0 and x=8. The cross-sections perpendicular to the axis on the interval 0
Answer:
The volume of the solid is 256 cubic units.
Step-by-step explanation:
Given:
The solid lies between planes [tex]x=0\ and\ x=8[/tex]
The cross section of the solid is a square with diagonal length equal to the distance between the parabolas [tex]y=-2\sqrt{x}\ and\ y=2\sqrt{x}[/tex].
The distance between the parabolas is given as:
[tex]D=2\sqrt x-(-2\sqrt x)\\\\D=2\sqrt x+2\sqrt x\\\\D=4\sqrt x[/tex]
Now, we know that, area of a square with diagonal 'D' is given as:
[tex]A=\frac{D^2}{2}[/tex]
Plug in [tex]D=4\sqrt x[/tex]. This gives,
[tex]A=\frac{(4\sqrt x)^2}{2}\\\\A=\frac{16x}{2}\\\\A=8x[/tex]
Now, volume of the solid is equal to the product of area of cross section and length [tex]dx[/tex]. So, we integrate it over the length from [tex]x=0\ to\ x=8[/tex]. This gives,
[tex]V=\int\limits^8_0 {A} \, dx\\\\V=\int\limits^8_0 {(8x)} \, dx\\\\V=8\int\limits^8_0 {(x)} \, dx\\\\V=8(\frac{x^2}{2})_{0}^{8}\\\\V=4[8^2-0]\\\\V=4\times 64\\\\V=256\ units^3[/tex]
Therefore, the volume of the solid is 256 cubic units.
This question is about volume calculation using calculus. The solid between two planes at x=0 and x=8 has cross-sections which, when described by a function of x A(x), the volume of the object can be computed via integration of A(x) dx from x=0 to x=8.
Explanation:The subject of this question falls under the field of Calculus, specifically, it's about Volume Calculation. The question describes a solid which is located between two planes at x=0 and x=8, perpendicular to the x-axis. Cross-sections perpendicular to the axis of this solid can be visualized like slices of the solid made along the x-axis.
If the area of these cross-sections can be represented by a function of x, A(x), then the volume of the entirety of the solid, V, can be calculated using the definite integral from x=0 to x=8 of A(x) dx. Essentially, this is summing up the volumes of the infinitesimal discs that make up the solid along the x-axis, from x=0 to x=8.
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