Answer:
The answer is 0.2865
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
In this problem, we have that:
The mean number of accidents on any given day in Coralville is 5. Of those, 25% are with an uninsured drive.
So [tex]\mu = 5*0.25 = 1.25[/tex]
Calculate the probability that on a given day in Coralville there are no trafficaccidents that involve an uninsured driver.
This is P(X = 0). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.25}*(1.25)^{0}}{(0)!} = 02865[/tex]
To calculate the probability of no traffic accidents involving uninsured drivers in Coralville, we can use the Poisson distribution formula with a mean of 5 accidents. The probability is approximately 0.00674, or 0.674%.
Explanation:To calculate the probability that on a given day in Coralville there are no traffic accidents that involve an uninsured driver, we can use the Poisson distribution formula. The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space, given the average rate of occurrence. In this case, the mean number of traffic accidents is 5, and the probability of an accident involving an uninsured driver is 0.25.
The formula for the Poisson distribution is P(X=k) = (e^(-λ) * λ^k) / k!, where λ is the mean number of events and k is the number of events we are interested in.
To find the probability of no traffic accidents involving uninsured drivers, we can plug in λ=5 and k=0 into the formula:
P(X=0) = (e^(-5) * 5^0) / 0!
P(X=0) = e^(-5) / 1 = 0.006737947
Therefore, the probability that on a given day in Coralville there are no traffic accidents that involve an uninsured driver is approximately 0.00674, or 0.674%.
A geologist manages a large museum collection of minerals, whose mass (in grams) is known to be normally distributed, with some mean µ and standard deviation σ. She knows that 60% of the minerals have mass less than a certain amount m, and needs to select a random sample of n = 16 specimens for an experiment. With what probability will their average mass be less than the same amount m?
Answer:
[tex]P(\bar X <m)=0.844[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the mass of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
And the z score is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
We know that for some amount m we have this:
[tex]P(X>m)=0.4[/tex] (a)
[tex]P(X<m)=0.6[/tex] (b)
We can use the z table or excel in order to find a quantile that satisfy the two conditions. The excel code would be:
"=NORM.INV(0.6,0,1)" and using condition (b) we have that Z=0.253
So we have this:
[tex]0.253=\frac{m-\mu}{\sigma}[/tex]
If we solve for m from the last equation we got:
[tex]m=0.253\sigma +\mu [/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
And we want this probability:
[tex]P(\bar X<m)[/tex]
We can apply the z score formula for this case given by:
[tex]z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{m-\mu}{\frac{\sigma}{\sqrt{16}}})[/tex]
[tex]P(Z<\frac{4(m-\mu)}{\sigma}[/tex]
If we use the expression obtained for m we got:
[tex]P(Z<\frac{4(0.253\sigma +\mu-\mu)}{\sigma}[/tex]
[tex]P(Z<\frac{1.012 \sigma}{\sigma})=P(Z<1.012)=0.844[/tex]
Abby bought two slices of pizza and
three bottles of water for $7.25
Cameron bought four slices of pizza and
one bottle of water for $8.25 what is the
solution
The solution is price of 1 slice of pizza is $ 1.75 and price of 1 bottle of water is $ 1.25
Solution:
Let "p" be the price of 1 slice of pizza
Let "b" be the price of 1 bottle of water
Given that Abby bought two slices of pizza and three bottles of water for $7.25
So we can frame a equation as:
two slices of pizza x price of 1 slice of pizza + three bottles of water x price of 1 bottle of water = $ 7.25
[tex]2 \times p + 3 \times b = 7.25[/tex]
2p + 3b = 7.25 ------- eqn 1
Cameron bought four slices of pizza and one bottle of water for $8.25
So we can frame a equation as:
four slices of pizza x price of 1 slice of pizza + one bottles of water x price of 1 bottle of water = $ 8.25
[tex]4 \times p + 1 \times b = 8.25[/tex]
4p + 1b = 8.25 ----- eqn 2
Let us solve eqn 1 and eqn 2 to find values of "p" and "b"
Multiply eqn 1 by 2
4p + 6b = 14.5 ----- eqn 3
Subtract eqn 2 from eqn 3
4p + 6b = 14.5
4p + 1b = 8.25
( - ) ------------------
5b = 6.25
b = 1.25Substitute b = 1.25 in eqn 1
2p + 3b = 7.25
2p + 3(1.25) = 7.25
2p + 3.75 = 7.25
2p = 3.5
p = 1.75Thus the solution is:
price of 1 slice of pizza = $ 1.75
price of 1 bottle of water = $ 1.25
The cost of one slice of pizza is $1.75. The cost of one bottle of water $1.25.
To find the cost of one slice of pizza and one bottle of water, we can set up a system of linear equations based on the information given:
Let p be the cost of one slice of pizza and w be the cost of one bottle of water.
1. Abby's purchase: 2p + 3w = 7.25
2. Cameron's purchase: 4p + 1w = 8.25
We can solve this system using either substitution or elimination. Let's use elimination.
Step 1: Eliminate \w
First, let's multiply the second equation by 3 to align the coefficients of w
3(4p + 1w) = 3(8.25)
12p + 3w = 24.75
Now our system of equations looks like this:
1. 2p + 3w = 7.25
2. 12p + 3w = 24.75
Step 2: Subtract the first equation from the second
12p + 3w) - (2p + 3w) = 24.75 - 7.25
10p = 17.50
Step 3: Solve for P
[tex]\[ p = \frac{17.50}{10} \][/tex]
p = 1.75
So, one slice of pizza costs $1.75
Step 4: Solve for w
Now substitute p = 1.75 back into the first equation:
2(1.75) + 3w = 7.25
w = 1.25
So, one bottle of water costs $1.25
The amounts of time that customers stay in a certain restaurant for lunch is normally distributed with a standard deviation of 17 minutes. A random sample of 50 lunch customers was taken at this restaurant. Construct a 99% confidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to two decimal places and use ascending order.
Answer:
99% CI: [45.60; 58.00]min
Step-by-step explanation:
Hello!
Your study variable is:
X: Time a customer stays in a certain restaurant. (min)
X~N(μ; σ²)
The population standard distribution is σ= 17 min
Sample n= 50
Sample mean X[bar]= 51.8 min
Sample standard deviation S= 27.68
You are asked to construct a 99% Confidence Interval. Since the variable has a normal distribution and the population variance is known, the statistic to use is the standard normal Z. The formula to construct the interval is:
X[bar] ± [tex]Z_{1-\alpha /2}[/tex]*(σ/√n)
[tex]Z_{1-\alpha /2} = Z_{0.995}= 2.58[/tex]
Upper level: 51.8 - 2.58*(17/√50) = 45.5972 ≅ 45.60 min
Lower level: 51.8 + 2.58*(17/√50) = 58.0027 ≅58.00 min
With a confidence level of 99%, you'd expect that the interval [45.60; 58.00]min will contain the true value of the average time customers spend in a certain restaurant.
I hope you have a SUPER day!
PS: Missing Data in the attached files.
Answer:
44.89 - 57.27
Exhibit 9-6a A random sample of 150 people was taken. 98 of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 57%. [R] Refer to Exhibit 9-6a. At the 0.1 level of significance, what conclusion do you draw?
(A) reject the null hypothesis
(B) fail to reject the null hypothesis
Answer:
Step-by-step explanation:
At the significance level given, the conclusion is B. Reject the null hypothesis.
What is the conclusion from the significance level?The first thing to do is to calculate the test statistic. This will be 1.98 from the z table.
In this case, the rejection region will be when the number is more than 1.28. Here, 1.98 is more than 1.28.
Therefore, the best thing to do is to simply reject the null hypothesis.
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A small entrepreneurial company is trying to decide between developing two different products that they believe they can sell to two potential companies, one large and one small. If they develop Product A, they have a 50% chance of selling it to the large company with annual purchases of about 20,000 units. If the large company won't purchase it, then they think they have an 80% chance of placing it with a smaller company, with sales of 15,000 units. On the other hand if they develop Product B, they feel they have a 40% chance of selling it to the large company, resulting in annual sales of about 17,000 units. If the large company doesn't buy it, they have a 50% chance of selling it to the small company with sales of 20,000 units.
1-What is the probability that Product B will being purchased by the smaller company?
A. 0.5
B. 0.4
C. 0.3
D. 0.8
Answer:
The correct option is C) 0.3
Step-by-step explanation:
Consider the provided information.
A small entrepreneurial can develop two product, product A and product B.
If they develop Product A:
There are 50% chance of selling it to the large company with annual purchases of about 20,000 units.
If the large company won't purchase it, then they think they have an 80% chance of placing it with a smaller company, with sales of 15,000 units.
If they develop Product B.
They have a 40% chance of selling it to the large company, resulting in annual sales of about 17,000 units. If the large company doesn't buy it, they have a 50% chance of selling it to the small company with sales of 20,000 units.
If they develop product B than the probability of purchasing by small company is 60% or 0.6
If large company doesn't buy it, then they have 50%or 0.5 chance of selling it to small company.
Hence, the total probability is: [tex]0.6\times0.5=0.3[/tex]
Therefore, the correct option is C) 0.3
Final answer:
The probability that Product B will be purchased by the smaller company is 30%, which corresponds to answer choice C.
Explanation:
The probability that Product B will be purchased by the smaller company is calculated by considering that Product B will only be offered to the smaller company if the large company does not purchase it. So, first, we must calculate the probability that the large company does not purchase Product B, which is 1 - 0.4 (since there is a 40% chance that the large company will purchase it). This results in a 60% chance that the large company will not purchase Product B. The probability that Product B will then be purchased by the smaller company is 50% of that 60%, which is calculated as 0.6 × 0.5 = 0.3 or 30%.
You are given the polar curve r = e^θ
List all of the points (r,θ) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of r and limit yourself to 1<_r<_1000 ( note the restriction on r) and 0≤θ<2????. If two or more points share the same value of r, list those starting with the smallest value of θ. If any blanks are unused, type an upper-case "N" in them.
Point 1: (r, θ) = ________
Point 2: (r, θ) = ________
Point 3: (r, θ) = ________
The tangent to [tex]r(\theta)=e^\theta[/tex] has slope [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], where
[tex]\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}[/tex]
By the chain rule, we have
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}[/tex]
and by the product rule,
[tex]\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta[/tex]
so that with [tex]\frac{\mathrm dr}{\mathrm d\theta}=e^\theta[/tex], we get
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}[/tex]
The tangent line is horizontal when the slope is 0; this happens for
[tex]-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi[/tex]
where [tex]n[/tex] is any integer. In the interval [tex]0\le\theta\le2\pi[/tex], this happens for [tex]n=1,2[/tex], or
[tex]\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4[/tex]
i.e at the points
[tex](r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)[/tex]
and
[tex](r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)[/tex]
Five students were tested before and after taking a class to improve their study habits. They were given articles to read which contained a known number of facts in each story. After the story each student listed as many facts as he/she could recall. The following data was recorded.
Before 10 12 14 16 12
After 15 14 17 17 20
The obtained value of the appropriate statistic is ____.
a. 3.92
b. 2.58
c. 4.12
d. 3.06
What do you conclude using a = 0.052 tail?
a. retain H0; we cannot conclude that the class improved study habits
b. accept H0; the class appeared to improve study habits
c. retain H0; the class had no effect on study habits
d. reject H0; the class appeared to improve study habits
Answer:
The obtained value of the appropriate statistic is ____.
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]
d. 3.06
[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]
d. reject H0; the class appeared to improve study habits
Step-by-step explanation:
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations we can use it.
Let put some notation
x=before method , y = after method
x: 10 12 14 16 12
y: 15 14 17 17 20
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]
The first step is define the difference [tex]d_i=y_i-x_i[/tex], that is given so we have:
d: 5,2,3,1,8
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=3.8[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =2.775[/tex]
The fourth step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=5-1=4[/tex]
Now we can calculate the p value, since we have a two tailed test the p value is given by:
[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]
The p value is less than the significance level given [tex]\alpha=0.05[/tex], so then we can conclude that we reject the null hypothesis.
d. reject H0; the class appeared to improve study habits
Help! How do you explain that √(m+n)=√m+ √n is not true for all values? (Example: m=5 and n=4)
Answer:
Square the expressions to see the difference.
Step-by-step explanation:
[tex]$ \sqrt{(m + n)} $[/tex].
Squaring this we have: [tex]$ (\sqrt{m + n})^2 = m + n $[/tex]
Now, [tex]$ \sqrt{m} + \sqrt{n} $[/tex]
Squaring this we get: [tex]$ (\sqrt{m})^2 + (\sqrt{n})^2 = m + n + 2 \sqrt{mn} $[/tex]
For the two expressions to be equal, we should have
[tex]$ m + n = m + n +2\sqrt{mn} $[/tex] ⇔ [tex]$ \sqrt{mn} = 0 $[/tex].
This is possible iff mn = 0. i.e, m = 0 or n = 0.
Otherwise, they are not equal.
When m = 5 and n = 4.
[tex]$ \sqrt{5 + 4} = \sqrt{9} = 3 $[/tex]
[tex]$ \sqrt{5} + \sqrt{4} = \sqrt{5} + 2 $[/tex]
First is an integer. Second is an irrational number.
Clearly, they are not equal.
Suppose that scores on the mathematics part of the National Assessment of Educational Progress (NAEP) test for eighth-grade students follow a Normal distribution with standard deviation σ = 110 . You want to estimate the mean score within ± 10 with 90 % confidence. How large an SRS of scores must you choose? Give your answer as a whole number.
Answer:
A sample size of at least 328 students is required.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, find the width M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
In this problem, we have that:
[tex]M = 10, \sigma = 110[/tex]
So:
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]10 = 1.645*\frac{110}{\sqrt{n}}[/tex]
[tex]10\sqrt{n} = 180.95[/tex]
[tex]\sqrt{n} = 18.095[/tex]
[tex]n = 327.43[/tex]
A sample size of at least 328 students is required.
Answer:
A sample size of at least 328 students is required.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find the width M as such
In which is the standard deviation of the population and n is the size of the sample.
In this problem, we have that:
So:
A sample size of at least 328 students is required.
This probability distribution shows the typical grade distribution fr a Geometry course with 35 students.
Using the probabilities given, find the probability that a student earns a grade of A.
p=
Enter a decimal rounded to the nearest hundredth.
Answer:
The probability that a student earns a grade of A is 1/7.
Step-by-step explanation:
Let E be an event and S be the sample space. The probability of E, denoted by P(E) could be computed as:
P(E) = n(E) / n(S)
As the total number of students = n(S) = 35
Students getting the grade A = n(E) = 5
So, the probability that a student earns a grade of A:
P(E) = n(E) / n(S)
= 5/35
= 1/7
Hence, the probability that a student earns a grade of A is 1/7.
Keywords: probability, sample space, event
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Answer:
.14
Step-by-step explanation:
The amounts of sugar (grams of sugar per gram of cereal) and calories (per gram of cereal) were recorded for a sample of 16 different cereals. The linear correlation coefficient is r= 0.765 and the regression equation is y ^ = 3.46 + 1.01 x , where x represents the amount of sugar. The mean of the 16 amounts of sugar is 0.295 grams and the mean of the 16 calorie counts is 3.76.
What is the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g?
Answer:
The best predicted calorie count for a cereal with a measured sugar amount of 0.40 g is 3.864
Step-by-step explanation:
Consider the provided information.
he linear correlation coefficient is r= 0.765 and the regression equation is [tex]\hat y= 3.46 + 1.01\times x[/tex]
The mean of the 16 amounts of sugar is 0.295 grams and the mean of the 16 calorie counts is 3.76.We need to find the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g
Substitute x = 0.40 in the given equation.
[tex]\hat y= 3.46 + 1.01\times 0.40[/tex]
[tex]\hat y= 3.46 + 0.404[/tex]
[tex]\hat y= 3.864[/tex]
Hence, the best predicted calorie count for a cereal with a measured sugar amount of 0.40 g is 3.864
When calculating the determinant of a matrix, the answer is a single number rather than a matrix.
Answer:
Yes, one of the properties of determinants is that they are real numbers (including zero) not matrix. This if the entries of the matrix are real. Determinants can be both positive or negative numbers.
Step-by-step explanation:
The determinant of a matrix is a number
What is a Matrix?
In mathematics, a matrix is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns, which is used to represent a mathematical object or a property of such an object
The determinant of inverse matrix can never be zero.
The determinant of an identity matrix is always 1.
The determinant of a zero matrix is always 0.
Given data ,
Let the matrix be represented as
[tex]A = \left[\begin{array}{ccc}a_{11} &a_{12} \\ a_{21} &a_{22} \\\end{array}\right][/tex]
Let the determinant of the matrix be d
Now , the determinant of the matrix A is calculated as
d = ( a₁₁ ) x ( a₂₂ ) - ( a₁₂ ) x ( a₂₁ )
So , the determinant will be a single number and will not be another matrix
The value of d will be scaling factor for the transformation of a matrix
Hence , the determinant of a matrix is a number
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The business college computing center wants to determine the proportion of business students who have personal computers (PC's) at home. If the proportion differs from 25%, then the lab will modify a proposed enlargement of its facilities. Suppose a hypothesis test is conducted and the test statistic is 2.4. Find the P-value for a two-tailed test of hypothesis.
Answer:
[tex]p_v =2*P(Z>2.4)=0.016[/tex]
And we can use the following excel code:
"=2*(1-NORM.DIST(2.4;0;1;TRUE))"
Step-by-step explanation:
1) Data given and notation
n represent the random sample taken
X represent the business students who have personal computers (PC's) at home
[tex]\hat p[/tex] estimated proportion of business students who have personal computers (PC's) at home
[tex]p_o=0.25[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is 0.25:
Null hypothesis:[tex]p=0.25[/tex]
Alternative hypothesis:[tex]p \neq 0.25[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
For this case the value of the statistic is given by z=2.4 and that's given.
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(Z>2.4)=0.016[/tex]
And we can use the following excel code:
"=2*(1-NORM.DIST(2.4;0;1;TRUE))"
You have a rather strange die: Three faces are marked with the letter A, two faces with the letter B, and one face with the letter C. You roll the die until you get a B. What is the probability that a B does not appear during the first three rolls?
Answer:
8/27 ≈ 29.6%
Step-by-step explanation:
Two of the six faces are B, which means four of the six faces are not B.
The probability of rolling not B three times is:
P = (4/6)^3
P = (2/3)^3
P = 8/27
P ≈ 29.6%
Answer:
8/27 = 0.296 = 29.6%
Step-by-step explanation:
Given that the die faces are as follows:
A A A B B C
i.e :
P( rolls A) = 3/6
P (rolls B) = 2/6
P (rolls C) = 1/6
for any single roll,
P (rolls not B) = P(rolls A) + P(rolls C)
P (rolls not B) = 3/6 + 1/6 = 4/6 = 2/3
for 3 consecutive rolls
P ( B does not appear) = P(rolls not B) * P(rolls not B) * P(rolls not B)
= P(rolls not B) ³
= (2/3)³
= 8/27 = 0.296 = 29.6%
A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 433 gram setting. It is believed that the machine is underfilling the bags.
A 26 bag sample had a mean of 427 grams with a standard deviation of 15.
Assume the population is normally distributed. A level of significance of 0.05 will be used.
Find the P-value of the test statistic.
You may write the P -value as a range using interval notation, or as a decimal value rounded to four decimal places.
Answer:
P-value = 0.0261
We conclude that the machine is under-filling the bags.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 433 gram
Sample mean, [tex]\bar{x}[/tex] = 427 grams
Sample size, n = 26
Alpha, α = 0.05
Sample standard deviation, σ = 15 grams
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 433\text{ grams}\\H_A: \mu < 433\text{ grams}[/tex]
We use one-tailed(left) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{427 - 433}{\frac{15}{\sqrt{26}} } = -2.0396[/tex]
Now, we calculate the p-value using the standard table.
P-value = 0.0261
Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.
We conclude that the machine is under-filling the bags.
The Pew Research Center Internet Project conducted a survey of 957 Internet users. This survey provided a variety of statistics on them. If required, round your answers to four decimal places. (a) The sample survey showed that 90% of respondents said the Internet has been a good thing for them personally. Develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally.
Answer: (0.881, 0.919)
Step-by-step explanation:
Let p be the population proportion of respondents who say the Internet has been a good thing for them personally.
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where z* = Critical value ,
[tex]\hat{p}[/tex] = Sample proportion
n = sample size.
As per given , we have
n= 957
[tex]\hat{p}=0.90[/tex]
By z-table , Critical value for 95% confidence interval : z*= 1.96
Then, the 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. will be :
[tex]0.90\pm 1.96\sqrt{\dfrac{0.90(1-0.90)}{957}}[/tex]
[tex]0.90\pm 1.96\sqrt{0.0000940438871473}[/tex]
[tex]0.90\pm 1.96(0.00969762275753)[/tex]
[tex]\approx0.90\pm 0.019\\\\=(0.90-0.019,\ 0.90+0.019)[/tex]
[tex](0.881,\ 0.919)[/tex]
Hence, the required 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally = (0.881, 0.919)
To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula: CI = p ± Z * sqrt((p * (1 - p)) / n). Plugging in the values, we can calculate a 95% confidence interval of [0.8808, 0.9192].
Explanation:To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula:
CI = p ± Z * sqrt((p * (1 - p)) / n)
Where:
CI is the confidence intervalp is the proportion of respondents who say the Internet has been a good thing for them (0.90)Z is the z-score corresponding to the desired confidence level (for 95% confidence level, Z ≈ 1.96)n is the sample size (957)Plugging in the values, we can calculate:
CI = 0.90 ± 1.96 * sqrt((0.90 * (1 - 0.90)) / 957)
This gives us a 95% confidence interval for the proportion, which is [0.8808, 0.9192]. So we can be 95% confident that the true proportion of respondents who say the Internet has been a good thing for them is between 0.8808 and 0.9192.
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The manager of a paint supply store wants to estimate the actual amount of paint contained in 1-gallon cans purchased from a nationally known manufacturer. The manufacturer's specifications state that the standard deviation of the amount of paint is equal to 0.02 gallon. A random sample of 50 cans is selected, and the sample mean amount of paint per 1-gallon can is 0.998 gallon. Complete parts (a) through (d).
a. Construct a 95% confidence interval estimate for the population mean amount of paint included in a 1-gallon can.
b. On the basis of these results, do you think the manager has a right to complain to the manufacturer? Why?
___because a 1-gallon paint can containing exactly 1-gallon of paint lies _____the 95% confidence interval.
c. Must you assume that the population amount of paint per can is normally distributed here? Explain.
d. Construct a 90% confidence interval estimate. How does this change your answer to part (b)?
How does this change your answer to part (b)?
A 1-gallon paint can containing exactly 1-gallon of paint lies ____the 90% confidence interval. The manager _______a right to complain to the manufacturer.
Answer:
a. 95% confidence interval estimate for the population mean amount of paint included in a 1-gallon can is 0.998±0.0055
b. No, because a 1-gallon paint can containing exactly 1-gallon of paint lies within the 95% confidence interval.
c. Yes. The population amount of paint per can is assumed normally distributed, because confidence interval calculations assume normal distribution of the parameter.
d. 90% confidence interval is 0.998±0.0046. The answer in b. didn't change; 1-gallon paint can containing exactly 1-gallon of paint lies within the 90% confidence interval. The manager doesn't have a right to complain to the manufacturer.
Step-by-step explanation:
Confidence Interval can be calculated using M±ME where
M is the sample mean amount of paint per 1-gallon can (0.998 gallon)
ME is the margin of error from the mean
And margin of error (ME) can be calculated using the equation
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the corresponding statistic in the 95% confidence level (1.96) s is the sample standard deviation (0.02 gallon)N is the sample size (50)Then ME=[tex]\frac{1.96*0.02}{\sqrt{50} }[/tex]≈0.0055
95% confidence interval is 0.998±0.0055
90% confidence interval can be calculated similary, only z statistic is 1.64.
ME=[tex]\frac{1.64*0.02}{\sqrt{50} }[/tex] ≈0.0046
90% confidence interval is 0.998±0.0046
Suppose ACT Composite scores are normally distributed with a mean of 21.2 and a standard deviation of 5.2. A university plans to offer tutoring jobs to students whose scores are in the top 1%.What is the minimum score required for the job offer? Round your answer to 1 decimal place.
Answer:
So the value of score that separates the bottom 99% of data from the top 1% is 33.316.
And rounded to 1 decimal place would be 33.3
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
2) Solution to the problem
Let X the random variable that represent the scores for the ACT of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(21.5,5.2)[/tex]
Where [tex]\mu=21.5[/tex] and [tex]\sigma=5.2[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.01[/tex] (a)
[tex]P(X<a)=0.99[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.99 of the area on the left and 0.01 of the area on the right it's z=2.33. On this case P(Z<2.33)=0.99 and P(Z>2.33)=0.01
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.99[/tex]
[tex]P(Z<\frac{a-\mu}{\sigma})=0.99[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]Z=2.33<\frac{a-21.2}{5.2}[/tex]
And if we solve for a we got
[tex]a=21.2 +2.33*5.2=33.316[/tex]
So the value of score that separates the bottom 99% of data from the top 1% is 33.316.
And rounded to 1 decimal place would be 33.3
Kim and Susan are playing a tennis match where the winner must win 2 sets in order to win the match.For each set the probability that Kim wins is 0.64. The probability of Kim winning the set is not affected by who has won any previous sets.(a) What is the probability that Kim wins the match?(b) What is the probability that Kim wins the match in exactly 2 sets (i.e. only 2 sets are played and Kim is the one who ends up winning)?(c) What is the probability that 3 sets are played?
Final answer:
To calculate the probability that Kim wins the match, we sum the probabilities of winning in 2 sets, 3 sets, and 4 sets. The probability that Kim wins the match is 0.6553. The probability that Kim wins the match in exactly 2 sets is 0.2339. The probability that 3 sets are played is 0.1728.
Explanation:
To calculate the probability that Kim wins the match, we need to consider the different possible outcomes. Kim can win the match in two sets, which happens with a probability of [tex](0.64)^2 = 0.4096[/tex] . Kim can also win the match in three sets, which happens with a probability of [tex](0.64)^2 * (1-0.64) * (1-0.64) = 0.1728[/tex]. And finally, Kim can win the match in four sets, which happens with a probability of [tex](0.64)^2 * (1-0.64)^2 = 0.0729.[/tex] Therefore, the probability that Kim wins the match is the sum of these probabilities: 0.4096 + 0.1728 + 0.0729 = 0.6553.
To calculate the probability that Kim wins the match in exactly 2 sets, we use the probability of winning a set twice and the probability of losing a set once:[tex](0.64)^2 * (1-0.64) = 0.2339.[/tex]
To calculate the probability that 3 sets are played, we need to consider the different possible outcomes. Kim can win the match in three sets, which happens with a probability of [tex](0.64)^2[/tex]* (1-0.64) * (1-0.64) = 0.1728. Therefore, the probability that 3 sets are played is 0.1728.
A village experienced 2% population growth, compounded continuously, each year for 10 years. At the end of the 10 years, the population was 158.
1. What was the population of the village at the beginning of the 10 years according to the exponential growth function? Round your answer up to the next whole number, and do not include units.
Answer:
The initial population at the beginning of the 10 years was 129.
Step-by-step explanation:
The population of the village may be modeled by the following function.
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the growth rate, in decimal.
In this problem, we have that:
[tex]P(10) = 158, r = 0.02[/tex].
So
[tex]158 = P_{0}e^{0.02*10}[/tex]
[tex]P_{0} = 158*e^{-0.2}[/tex]
[tex]P_{0} = 129[/tex]
The initial population at the beginning of the 10 years was 129.
The population of the village at the beginning of the 10 years was approximately 130. This is determined by using the formula for exponential growth that is continuously compounded and solving for the initial population.
Explanation:We can use the formula for exponential growth that is continuously compounded, which is P=Pe^rt, where P is the final population, P is the initial population, r is the growth rate, and t is the time.
In this case, we know that P=158, r=0.02 (2% expressed as a decimal), and t=10. We need to find P.
Substituting the given values into the formula, we get 158=Pe^(0.02*10).
We can calculate e^(0.02*10) using a calculator to get approximately 1.22. Therefore, the equation becomes 158=P*1.22.
To find P, we can divide both sides of the equation by 1.22 resulting in P ≈ 158/1.22 ≈ 129.5. Since we are asked to round up to the next whole number, the original population was approximately 130.
Learn more about Exponential GrowthThe marginal average cost of producing x digital sports watches is given by the function C'^overbar(x), where C^overbar(x) is the average cost in dollars. C'^overbar(x) = 1, 200/x^2, C^overbar(100) = 22 Find the average cost function and the cost function. What are the fixes costs? The average cost function is C^overbar(x) = ____. The cost function m C(x) = ____. The fixed costs are $____.
Answer:
The average cost function is [tex]\bar C(x)=-\frac{1200}{x}+34[/tex].
The cost function is [tex]C(x)=-1200+34x[/tex].
The fixed costs are -$1200.
Step-by-step explanation:
If [tex]y = C(x)[/tex] is the total cost of producing x items, then the average cost, or cost per unit, is [tex]\bar C(x)=\frac{C(x)}{x}[/tex] and the marginal average cost is [tex]\bar C'(x)=\frac{d}{dx}\bar C(x)[/tex].
We know that the marginal average cost is given by
[tex]\bar C'(x)=\frac{1200}{x^2}[/tex]
To find the average cost function, we find the indefinite integral of [tex]\frac{1200}{x^2}[/tex] and determine the arbitrary integration constant using [tex]\bar C(100)=22[/tex][tex]\bar C'(x)=\frac{1200}{x^2}\\\\\bar C(x)=\int {\frac{1200}{x^2}} \, dx[/tex]
[tex]\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\1200\cdot \int \frac{1}{x^2}dx\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\1200\cdot \frac{x^{-2+1}}{-2+1}\\\\-\frac{1200}{x}\\\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\\\-\frac{1200}{x}+C[/tex]
Because [tex]\bar C(100)=22[/tex] C is
[tex]22=-\frac{1200}{100}+C\\C=34[/tex]
And the average cost function is
[tex]\bar C(x)=-\frac{1200}{x}+34[/tex]
To find the cost function we multiply by x the average cost function according with the above definition[tex]x\cdot \bar C(x)=C(x)\\\\C(x)=(-\frac{1200}{x}+34)\cdot x\\\\C(x)=-1200+34x[/tex]
Cost functions are express as
[tex]C=(fixed \:costs)+(variable \:costs)\\C=a+bx[/tex]
According with this definition and the cost function that we have the fixed costs are -$1200.
To find the average cost function, use the given formula and substitute the value of x. To find the cost function, integrate the average cost function. The fixed costs are $22.
Explanation:To find the average cost function, we will substitute the value of C^overbar(x) into the formula. The average cost function is given by C^overbar(x) = 1,200/x^2.
To find the cost function, we will integrate the average cost function. The cost function m C(x) is the integral of C^overbar(x) with respect to x. F
inally, to find the fixed costs, we will substitute the value of C^overbar(100) into the average cost function.
The average cost function is C^overbar(x) = 1,200/x^2.
The cost function m C(x) = 1,200/x.
The fixed costs are $22.
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Question 5. We have 5-year statistics of average amount of wheat crop (tons) harvested from 1 km2 per year, the results are as follows: 560, 525, 496, 543, 499. Test the hypothesis that the mean wheat crop is 550 tons per 1km2 (α = 0.05) and choose the correct answer. Determine a 95% confidence interval on the mean wheat crop. Determine whether the hypothesis that the mean crop is 550 tons p
Answer:
a) [tex]t=\frac{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052[/tex]
[tex]p_v =P(t_{(4)}<-2.052)=0.055[/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 FAIL to reject the null hypothesis.
We can say that at 5% of significance the true mean is not significantly different from 550.
b) The 95% confidence interval would be given (490.184;559.016).
We are confident at 95% that the true mean is between (490.184;559.016).
Step-by-step explanation:
Part a
Data given and notation
560, 525, 496, 543, 499
The mean and sample deviation can be calculated from the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n x_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}[/tex]
[tex]\bar X=524.6[/tex] represent the sample mean
[tex]s=27.682[/tex] represent the sample standard deviation
[tex]n=5[/tex] sample size
[tex]\mu_o =550[/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 population mean is equal to 550, the system of hypothesis are:
Null hypothesis:[tex]\mu = 550[/tex]
Alternative hypothesis:[tex]\mu \neq 550[/tex]
Since we don't know the population deviation, 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{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052[/tex]
P-value
We need to calculate the degrees of freedom first given by:
[tex]df=n-1=5-1=4[/tex]
Since is a one-side left tailed test the p value would given by:
[tex]p_v =P(t_{(4)}<-2.052)=0.055[/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 FAIL to reject the null hypothesis.
We can say that at 5% of significance the true mean is not significantly different from 550.
Part b
The confidence interval for a proportion is given by this formula
[tex]\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
The degrees of freedom are 4 from part a
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]t_{\alpha/2}=2.78[/tex]
And replacing into the confidence interval formula we got:
[tex]524.6 - 2.78 \frac{27.682}{\sqrt{5}}=490.184[/tex]
[tex]524.6 + 2.78 \frac{27.682}{\sqrt{5}}=559.016[/tex]
And the 95% confidence interval would be given (490.184;559.016).
We are confident at 95% that the true mean is between (490.184;559.016).
The commercial for the new Meat Man Barbecue claims that it takes 10 minutes for assembly. A consumer advocate thinks that the claim is false. The advocate surveyed 50 randomly selected people who purchased the Meat Man Barbecue and found that their average time was 11.2 minutes. The standard deviation for this survey group was 3.1 minutes. What can be concluded at the 0.05 level of significance?H0: mu.gif= 10Ha: mu.gif [ Select ] [">", "<", "Not Equal to"] 10Test statistic: [ Select ] ["Chi-square", "F", "Z", "t"] p-Value = [ Select ] ["0.105", "0.009", "0.091", "0.025"] . Round your answer to three decimal places. [ Select ] ["Reject the null hypothesis", "Fail to reject the null hypothesis"] Conclusion: There is [ Select ] ["sufficient", "insufficient"] evidence to make the conclusion that the population mean amount of time to assemble the Meat Man barbecue is not equal to 10 minutes.
Answer:
There is enough evidence to make the conclusion that the population mean amount of time to assemble the Meat Man barbecue is not equal to 10 minutes (P-value=0.009).
Step-by-step explanation:
We have to perform an hypothesis test on the mean.
The null and alternative hypothesis are:
[tex]H_0: \mu=10\\\\H_1: \mu \neq 10[/tex]
The significance level is [tex]\alpha=0.05[/tex].
The test statistic t can be calculated as:
[tex]t=\frac{M-\mu}{s/\sqrt{N} } =\frac{11.2-10}{3.1/\sqrt{50} }=2.737[/tex]
The degrees of freedom are:
[tex]df=N-1=50-1=49[/tex]
The P-value (two-tailed test) for t=2.737 and df=49 is P=0.00862.
This P-value (0.009) is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.
There is enough evidence to make the conclusion that the population mean amount of time to assemble the Meat Man barbecue is not equal to 10 minutes.
Final answer:
A hypothesis test can be conducted to test the claim made in the commercial for the new Meat Man Barbecue. The test results indicate that there is evidence to support that the population mean assembly time is not equal to 10 minutes.
Explanation:
To test the claim made in the commercial for the new Meat Man Barbecue, a hypothesis test can be conducted. The null hypothesis, H0, states that the population mean assembly time is 10 minutes, while the alternative hypothesis, Ha, states that it is not equal to 10 minutes. The test statistic to use in this case is the t-test, as we are comparing the sample mean to a known value. The p-value for this test is 0.009, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis and conclude that there is evidence to support that the population mean assembly time is not equal to 10 minutes.
Show that the class of closed intervals [a, b], where a is rational and b is irrational and a < b, is a base for a topology on the real line R.
Answer:
See proof below.
Step-by-step explanation:
Remember that a collection of subsets B of a set X is a base for a topology of X if the following conditions hold
The union of B is equal to XIf T,S∈B and p∈T∩S then there exists some Q∈B such that x∈Q and Q⊆T∩S.In this case, X=R and B={[a,b]: a is rational and b is irrational}. Let's prove the statements above
The inclusion ∪B⊆R is true for any collection of subsets of the real line. To prove that R⊆∪B., let x∈R. If x is rational choose an irrational number y>x. (this can be done because irrational numbers are not bounded above) Then x∈[x,y] and [x,y]∈B, thus x∈∪B. If x is irrational, choose some rational number z<x (it can be done because rationals are not bounded below). Then x∈[z,x] for some [z,x]∈B, thus by definition of union, x∈∪B. In any case, x∈∪B therefore R=∪B.Let T=[a,b], S=[c,d]∈B be arbitrary elements of B. Suppose that p∈T∩S. Define Q=T∩S. Q is a closed interval, its starting point is either a or c (the greatest of these), which are rational, and its endpoint is either b or d (the smallest of these), which are irrational. In any case Q∈B and we have that p∈Q⊆T∩S.A professor expects that the grades on a recent exam are normally distributed with a mean of 80 and a standard deviation of 10. She records the actual distribution of grades and wants to compare it to the normal distribution.
Which of the following χ2 tests should be used in the situation above?
Select the correct answer below:
A. Test of Independence
B. Goodness-of-Fit Test
C. Test for Homogeneity
Answer:
B. Goodness-of-Fit Test
Step-by-step explanation:
Given that a professor expects that the grades on a recent exam are normally distributed with a mean of 80 and a standard deviation of 10.
She records the actual distribution of grades and wants to compare it to the normal distribution
For this expected values are to be got assuming normal and the observed and expected are used to calculate chi square.
Depending on the chi square statistic conclusion is made
Here the test to be done is
B) Goodness of fit test.
A is wrong because test of independence is done when there are more than one categorical variable in the rows.
C is wrong because here homogeneity is not tested
Only B is right.
A sample of 106 healthy adults have a mean body temperature of 98.2, and a standard deviation of 0.62. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. Group of answer choices ( 98.04, 98.36 ) ( 97.54, 97.95 ) ( 97.84, 98.12 ) ( 97.82, 98. 15 ) ( 97.95, 98.15 )
Answer: [tex](98.04,\ 98.36)[/tex]
Step-by-step explanation:
Given : Sample size of healthy adults: n= 106
Degree of freedom = df =n-1 = 105
Sample mean body temperature : [tex]\overline{x}=98.2[/tex]
Sample standard deviation : [tex]s= 0.62[/tex]
Significance level ; [tex]\alpha= 1-0.99=0.01[/tex]
∵ population standard deviation is unknown , so we use t- critical value.
Confidence interval for the population mean :
[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
Using t-distribution table , we have
Critical value for df = 105 and [tex]\alpha=0.01[/tex]= [tex]t_{\alpha/2, df}=t_{0.005 , 105}=2.623[/tex]
A 99% confidence interval estimate of the mean body temperature of all healthy humans will be :
[tex]98.2\pm (2.623)\dfrac{0.62}{\sqrt{106}}[/tex]
[tex]98.2\pm (2.623)(0.0602197234662)[/tex]
[tex]98.2\pm (0.157956334652)[/tex]
[tex]\approx98.2\pm 0.16[/tex]
[tex](98.2-0.16,\ 98.2+0.16)[/tex]
[tex](98.04,\ 98.36)[/tex]
Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans = [tex](98.04,\ 98.36)[/tex]
Use Cramer's Rule to find x in the system of equations below.
2x − 3y = 17
5x + 4y = 8
Answer: x = 4 , y = -3
Step-by-step explanation:
Going by the Cramer's rule , we first determine the determinant by dealing with only the coefficients of x and y in the 2 x 2 matrix.
2x - 3y = 17
5x + 4y = 8
2 -3
5 4, going by the rule now, we now have
(2 x 4) - (5 x -3)
8 + 15
= 23.
Now to find the value of x , replace the constants with the coefficient of x and divide by he determinants.
17 -3
8 4
---------------
2 -3
5 4
( 17 x 4 ) - ( 8 x -3 )
---------------------------
23
= 68 + 24
------------
23
= 92/23
= 4.
x = 4
Now to find y, just repeat the process by replacing the coefficient of y with the constants.
2 17
5 8
-----------
2 -3
5 4
( 2 x 8 ) - ( 5 x 17 )
-----------------------
23
16 - 85
---------
23
= -69/23
= -3
y = -3.
check
substitute for the values in any of the equations above.
2(4) - 3(-3)
8 + 9
= 17
which of the following number sets does 82 squares belong in?
1,2,3 and 4
1 and 3
1 and 2
2, 3, and 4
The square root of 81 repeating is 9.05, (Rounded)
-It goes on forever.
Since it goes on forever, it's not a whole number or a natural number, both kind of numbers never keep going on forever.
Which leaves us with Irrational and Real Numbers
Therefore, 1 and 2 is your answer.
Best of Luck to you.
If you have any questions, or need more info, feel free to comment below.
Happy New Year!
An international polling agency estimates that 36 percent of adults from Country X were first married between the ages of 18 and 32, and 26 percent of adults from Country Y were first married between the ages of 18 and 32. Based on the estimates, which of the following is closest to the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y (Country X minus Country Y) who were first married between the ages of 18 and 32 is greater than 0.15?
(A) 0.1398
(B) 0.2843
(C) 0.4315
(D) 0.5685
(E) 0.7157
Answer:
(B) 0.2843
Step-by-step explanation:
Let d be the difference in proportions from Country X and Country Y who were first married between the ages of 18 and 32.
Then hypotheses are
[tex]H_{0}[/tex]: d=0.15
[tex]H_{a}[/tex]: d<0.15
Test statistic can be found using the equation
[tex]z=\frac{p1-p2-0.15}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where
p1 is the sample proportion of Country X (0.36)p2 is the sample proportion of Country Y (0.26)p is the pool proportion of p1 and p2 ([tex]\frac{60*0.36+50*0.26}{50+60}=0.31[/tex])n1 is the sample size of adults from Country X (60)n2 is the sample size of adults from Country Y (50)Then [tex]z=\frac{0.36-0.26-0.15}{\sqrt{{0.31*0.69*(\frac{1}{60} +\frac{1}{50}) }}}[/tex] ≈ 0.5646
p-value of test statistic is ≈ 0.2843
p-value states the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is at least 0.15
To find the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is greater than 0.15, we need to calculate the sampling distribution of the difference in proportions. The answer is (E) 0.7157.
Explanation:To find the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is greater than 0.15, we need to calculate the sampling distribution of the difference in proportions.
The standard error of the difference in proportions can be calculated as:
SE = sqrt((p1 * (1 - p1)) / n1 + (p2 * (1 - p2)) / n2)
where p1 and p2 are the proportions of first married adults from Country X and Country Y respectively, and n1 and n2 are the sample sizes for Country X and Country Y.
Let's plug in the values:
SE = sqrt((0.36 * (1 - 0.36)) / 60 + (0.26 * (1 - 0.26)) / 50)
SE = sqrt(0.0024 / 60 + 0.0016 / 50)
SE = sqrt(0.00004 + 0.00003)
SE = sqrt(0.00007)
SE ≈ 0.00836660027
Now, we can use the standard normal distribution to find the probability that the difference in proportions is greater than 0.15 by calculating the z-score and looking it up in the z-table or using a calculator.
Z = (0.15 - 0) / 0.00836660027
Z ≈ 17.914253
The probability that the difference in proportions is greater than 0.15 is very close to 1, which means it is highly likely that the difference in proportions will be greater than 0.15 in a random sample. Therefore, the answer is (E) 0.7157.
Learn more about Sampling Distribution of Difference in Proportions here:https://brainly.com/question/32127904
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A coin is tossed 3 times. What is the probability that the number of tails obtained will be between 1 and 3 inclusive?
Answer: [tex]\dfrac{7}{8}[/tex]
Step-by-step explanation:
Given : When a coin is tossed twice , the total outcomes will become
HHH , HHT , HTH , THH , TTH , THT , HTT ,TTT
Total outcomes = 8
Favorable outcomes to obtain 1 and 3 inclusive = HHT , HTH , THH , TTH , THT , HTT ,TTT
Number of favorable outcomes = 7
Probability that the number of tails obtained will be between 1 and 3 inclusive = [tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
[tex]=\dfrac{7}{8}[/tex]
Hence, the probability that the number of tails obtained will be between 1 and 3 inclusive [tex]=\dfrac{7}{8}[/tex]