Answer:
±1, ±2, ±7, ±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
Find the expected value of the following probability distribution. X 1 2 3 4 5 P(X) 0.05 0.20 0.35 0.25 0.15 Round your answer to two decimal place as necessary. For example, 4.56 would be a legitimate entry.Expected value = ___.
Answer:
5.63
Step-by-step explanation:
The expected value of the given probability distribution is calculated by multiplying each value by its probability and summing the results, yielding an expected value of 3.25.
To find the expected value of the given probability distribution, you multiply each value of X by its corresponding probability P(X) and then sum those products. Mathematically, the expected value E(X) is calculated as:
E(X) = 1 * 0.05 + 2 * 0.20 + 3 * 0.35 + 4 * 0.25 + 5 * 0.15E(X) = 0.05 + 0.40 + 1.05 + 1.00 + 0.75E(X) = 3.25The expected value is 3.25, rounded to two decimal places as instructed.
1. An equation is shown below.
3 (x-2) + 7x= 1/2(6x-2)
How many solutions, if any, does the equation have?
Answer:
x=5/7
Step-by-step explanation:
3(x-2)+7×=1/2×(6×-2)
3x-6+7×=1/2×2(3×-1)
3×-6+7×=3×-1
-6+7×=-1
7×=-1+6
7×=5
A test statistic is found to have a P-value of .015. Which of the following statements are true?
I. This finding is significant at the .05 level of significance.
II. This finding is significant at the .01 level of significance.
III. The probability of getting a test statistic as extreme (or more extreme) as this one by chance alone is .015.
a. I and III only
b. I only
c. II only
d. III only
e. I and II only
Answer:
Option a) I and III only
Step-by-step explanation:
given that a test statistic is found to have a P-value of .015.
In hypothesis testing after finding out test statistic as difference between hypothesises mean and sample mean divided by standard error of sample we find out p value.
Significance level, known as alpha is fixed as 5% or 1% or 10% depending on the matter of interest
If p value calculated is less than our significance level i.e. alpha we reject null hypothesis
Here p = 0.015 is less than 5% and 10% but greater than 1%
So we reject at 5% and 10% level and but accept at 1% level
Correct options would be:
I. This finding is significant at the .05 level of significance.
III. The probability of getting a test statistic as extreme (or more extreme) as this one by chance alone is .015
Option a) I and III only
Option a is correct that statements I and III are correct.
The P-value of .015 indicates the probability of obtaining a test statistic as extreme as the one observed if the null hypothesis is true. When interpreting this:
i. This finding is significant at the .05 level of significance: Since .015 < .05, we reject the null hypothesis at the 5% level.
ii. This finding is not significant at the .01 level of significance: Since .015 > .01, we do not reject the null hypothesis at the 1% level.
iii. This statement correctly interprets the meaning of the P-value. The P-value represents the probability of obtaining a test statistic as extreme or more extreme than the observed one under the null hypothesis. This statement is true.
Therefore, statements I and III are true, making the correct answer a. I and III only.
The 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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How many integers between 360 and 630 are there such that they have odd number of divisors?
Answer:
7
Step-by-step explanation:
Concept to Know: "A number is a perfect square if and only if it has odd number of positive divisors "
Find all the squared values that lies between 360 and 630
360< 19², 20², 21², 22², 23², 24², 25² < 630
19², 20², 21², 22², 23², 24², and 25² are all the squared values that lies between 360 and 630. There are seven of those squared numbers so the answer is 7.
There are 7 integers between 360 and 630 that are perfect squares, hence have odd numbers of divisors. These numbers are the squares of the integers from 19 to 25.
Explanation:This is a problem in the field of number theory. In mathematics, only perfect square numbers have an odd number of total divisors. This is because factors or divisors usually come in pairs, but in perfect squares, one pair is actually a duplicate (the square root of the number itself). Hence, we need to find the perfect squares between 360 and 630.
By performing square root operation, we find that squares of integers from 19 to 25 are in the range given. So, the numbers are 361 (192), 400 (202), 441 (212), 484 (222), 529 (232), 576 (242) and 625 (252).
Therefore, there are 7 integers between 360 and 630 that have an odd number of divisors.
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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!
Laser scanning for fish volume estimation. engineers design tanks for rearing commercial fish to minimize both the use of natural resources (water) and the rearing volume necessary to ensure fish welfare.
One key to a well-designed tank is obtaining a reliable estimate of the volume (biomass) of fish reared in the tank.
The feasibility of a laser scanning technique for estimating fish biomass was investigated in the Journal of Aquacultural Engineering (Nov.2012). Fifty turbot fish were reared in a tank for experimental purposes.
A laser scan was executed in four randomly selected locations in the tank and the volume (in kilograms) of fish layer at each location was measured.
The four laser scans yielded a mean volume of 240 kg with a standard deviation of 15 kg. from this information, estimate the true mean volume of fish layer in the tank with 99% confidence. Interpret the result, practically.
What assumption about the data is necessary for the inference derived from the analysis to be valid?
Answer:
The 99% confidence interval would be given by (196.2;283.8)
We are 99% condident that the true mean is between 196.2 and 283.8
We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
[tex]\bar X=240[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=15 represent the sample standard deviation
n=4 represent the sample size
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=4-1=3[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,3)".And we see that [tex]t_{\alpha/2}=5.84[/tex]
Now we have everything in order to replace into formula (1):
[tex]240-5.84\frac{15}{\sqrt{4}}=196.2[/tex]
[tex]240+5.84\frac{15}{\sqrt{4}}=283.8[/tex]
So on this case the 99% confidence interval would be given by (196.2;283.8)
We are 99% condident that the true mean is between 196.2 and 283.8
What assumption about the data is necessary for the inference derived from the analysis to be valid?
We need to assume that the data comes from a random sample and we need to assume that the distribution of the data is normal.
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
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.
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%
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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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 radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5551 years. Suppose C(t) is the amount of carbon-14 present at time t.
(a) Find the value of the constant kk in the differential equation C′=−kC.
Answer:
k-0.000125
Step-by-step explanation:
Given that the radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5551 years.
[tex]C' = -kC\\\frac{dC}{C} =-kdt\\ln C = -kt+C_1[/tex], where C_1 is arbitrary constant.
Or [tex]C(t) = Ae ^{-kt}[/tex]
To find K.
C(t) = 1/2 C when t = 5551
i.e. A will become A/2 in 5551 years
[tex]A/2 = Ae^{-5551k} \\ln 0.5= -5551k\\k = 0.000125[/tex]
It is known that roughly 2/3 of all human beings have a dominant right foot or eye. Is there also right-sided dominance in kissing behavior? An article reported that in a random sample of 130 kissing couples, both people in 84 of the couples tended to lean more to the right than to the left. (Use α = 0.05.) If 2/3 of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 130 who do so differs from the expected value by at least as much as what was actually observed? (Round your answer to three decimal places.)
The probability that the number in a sample of 130 who do so differs from the expected value is 0.6198.
How to calculate probability?From the information, the critical value is given as 84. Therefore, the mean will be:
= np = 130 × 2/3 = 86.67
The standard deviation is also 5.3748. The corresponding z score will be:
= (84 - 86.67)/5.3748
= -0.50.
Therefore, the left tailed area will be:
P(z < -0.50) = 0.3099
Since, it's two tailed, we'll multiply by 2. This will be:
= 2 × 0.3099
= 0.6198
In conclusion, the probability is 0.6198.
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The probability that the number differs from the expected value by at least as much is approximately 0.620 (rounded to three decimal places).
To determine the probability that the number in a sample of 130 kissing couples who lean to the right differs from the expected value by at least as much as observed, we will conduct a hypothesis test using the binomial distribution. We can then approximate this with a normal distribution since the sample size is large.
Given:
- Sample size [tex](\( n \)) = 130[/tex]
- Observed number of right-leaning couples [tex](\( X \)) = 84[/tex]
- Expected proportion of right-leaning couples [tex](\( p \)) = \(\frac{2}{3}\)[/tex]
- Significance level [tex](\( \alpha \))[/tex] = 0.05
Step 1: Calculate the expected number and standard deviation
The expected number of right-leaning couples is:
[tex]\[ \mu = n \cdot p = 130 \cdot \frac{2}{3} \approx 86.67 \][/tex]
The standard deviation is:
[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{130 \cdot \frac{2}{3} \cdot \frac{1}{3}} \approx \sqrt{130 \cdot \frac{2}{9}} \approx \sqrt{28.89} \approx 5.375 \][/tex]
Step 2: Calculate the z-score for the observed value
The z-score measures how many standard deviations the observed value is away from the expected value:
[tex]\[ z = \frac{X - \mu}{\sigma} = \frac{84 - 86.67}{5.375} \approx \frac{-2.67}{5.375} \approx -0.497 \][/tex]
Since we are interested in the probability that the number differs from the expected value by at least as much as observed, we need to consider both tails of the distribution.
Step 3: Calculate the probability using the standard normal distribution
The probability for a z-score of [tex]\( \pm 0.497 \)[/tex] can be found using the standard normal distribution table or a calculator. For z = 0.497 :
[tex]\[ P(Z \leq 0.497) \approx 0.690 \][/tex]
Since we are considering both tails:
[tex]\[ P(|Z| \geq 0.497) = 2 \cdot (1 - 0.690) = 2 \cdot 0.310 = 0.620 \][/tex]
Step 4: Conclusion
The probability that the number of right-leaning couples in a sample of 130 differs from the expected value by at least as much as observed is approximately 0.620.
So, the rounded answer to three decimal places is:
[tex]\[ \boxed{0.620} \][/tex]
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.
The width of a rectangle is 6 inches shorter than 3 times it’s length. The width of the rectangle is at least 45 millimeters. Write and solve an inequality to determine the possible length, x, of the rectangle
An inequality to determine the possible length, x, of the rectangle is 3x - 6 ≥ 45/25.4.
The possible length, x is 2.5906 inches.
How to calculate the area of a rectangle?In Mathematics and Geometry, the area of a rectangle can be calculated by using the following mathematical equation:
A = xy
Where:
A represent the area of a rectangle.y represent the width of a rectangle.x represent the length of a rectangle.Conversion:
25.4 millimeters = 1 inches
45 millimeters = 45/25.4 inches.
Since the width of this rectangle is 6 inches shorter than 3 times it’s length and the width of the rectangle is at least 45 millimeters, an inequality to determine the possible length, x, of the rectangle can be written as follows;
3x - 6 ≥ 45/25.4
3x - 6 ≥ 1.7717
3x ≥ 1.7717 + 6
3x ≥ 7.7717
x ≥ 7.7717/3
x ≥ 2.5906 inches.
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Suppose there are two types of persons: high-ability and low-ability. A particular diploma costs a high-ability person $10,000 and costs a low-ability person $18,000. Firms wish to use education as a screening device where they intend to pay $35,000 to workers without a diploma and $K to those with a diploma. In what range must K be to make this an effective screening device?
Answer:
K must be in range $45000 ≤ k ≤ $53000 to make this an effective screening device.
Step-by-step explanation:
In order for the screening device to be effective, the range of k must be such that, it is more than the pay of a worker without diploma. The extra amount must at least be equal to the cost of diploma
So the value of k must be in range from (35000 + 10000) to (35000 + 18000)
Hence,
$45000 ≤ k ≤ $53000
The range of wages paid to workers with a diploma that makes education an effective screening device is $18,001 to $34,999.
Explanation:In order for education to be an effective screening device for firms, the wage paid to workers with a diploma should be higher than the wage paid to workers without a diploma. Let's consider the costs: a high-ability person pays $10,000 for a diploma, while a low-ability person pays $18,000. The wage paid to workers without a diploma is $35,000. The range of K (the wage paid to workers with a diploma) that makes education an effective screening device is $18,001 to $34,999.
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))"
A random sample of the correct choice on 400 multiple-choice questions on a variety of AP exams1 shows that B was the most common correct choice, with 90 of the 400 questions having B as the answer. Does this provide evidence that B is more likely to be the correct choice than would be expected if all five options were equally likely? Show all details of the test. The data are available in APMultipleChoice.
a) State the null and alternative hypotheses
b) Calculate the test statistic and p-value
Answer:
a) Null hypothesis:[tex]p\leq 0.2[/tex]
Alternative hypothesis:[tex]p > 0.2[/tex]
b) [tex]z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25[/tex]
[tex]p_v =P(Z>1.25)=0.106[/tex]
Step-by-step explanation:
1) Data given and notation
n=400 represent the random sample taken
X=90 represent the number of questions with B as the correct answer
[tex]\hat p=\frac{90}{400}=0.225[/tex] estimated proportion of arrests that were not prosecuted
[tex]p_o=0.2[/tex] is the value that we want to test, since we assume that each question present 5 options and just one is correct, 1/5 =0.2 if all five options were equally likely
[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 higher than 0.2.:
Null hypothesis:[tex]p\leq 0.2[/tex]
Alternative hypothesis:[tex]p > 0.2[/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
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.225 -0.2}{\sqrt{\frac{0.2(1-0.2)}{400}}}=1.25[/tex]
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 significance level provided [tex]\alpha[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>1.25)=0.106[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of B correct answers is not significantly higher than 0.2.
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.
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The scale drawing has a scale of 1/2 in: 8 mi. Find the length on the drawing for 2 in Please answer asap
Answer:the length on the drawing for 2 in is 8 miles
Step-by-step explanation:
The scale drawing has a scale of 1/2 in: 8 miles. This means that every 1/2 inch on the drawing represents an actual length of 2 miles.
Let us determine the actual length that 1 inch on the drawing will represent.
1 inch would represent 2/(1/2) = 4 miles.
Therefore, the actual length that 2 inches on the drawing represents would be 4 × 2 = 8 miles
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.
Collection of_______ is called Statistics.
a. Methods
b. Sample data
c. Numerical information
d. Population data
e. Cleaned data
Answer:
Collection of numerical information is called Statistics
Step-by-step explanation:
From the Oxford dictionary, the definition of statistics (science) is: "the practice or science of collecting and analyzing numerical data in large quantities, especially to infer proportions in a whole from those in a representative sample".
While the definition of statistic (data) is: "a fact or piece of data obtained from a study of a large quantity of numerical data".
Statistics is a science, not a method, therefore answer can´t be a).
Not all sample data is necessarily a statistic, therefore answer can´t be b).
All statistics are numerical information.
Not all statistics are population data, therefore answer can´t be d).
Cleaned data is information filtered with a specific criterion, but not necessarily used in a statistic.
The most probable answer is c.
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]
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 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.
At which three points x1, x2, and x3 closest to x=0 but with x>0 will the displacement of the string y(x,t) be zero for all times? These are the first three nodal points. Express the first three nonzero nodal points as multiples of the wavelength λ, using constants like π. List the factors that multiply λ in increasing order, separated by commas.
The first three nonzero nodal points are at[tex]\( \frac{\lambda}{2}, \lambda, \frac{3\lambda}{2} \),[/tex] where the factors multiplying [tex]\( \lambda \)[/tex] are [tex]\( \frac{1}{2}, 1, \frac{3}{2} \).[/tex]
To find the first three nodal points of the displacement of the string [tex]\( y(x, t) \)[/tex], where x > 0, we need to consider the standing wave pattern formed on the string. In a standing wave, the displacement is zero at certain points along the string for all times.
For a standing wave on a string fixed at both ends, the displacement y(x, t) can be expressed as:
[tex]\[ y(x, t) = A \sin(kx) \cos(\omega t) \][/tex]
Where:
- A is the amplitude of the wave.
- k is the wave number.
- [tex]\( \omega \)[/tex] is the angular frequency.
The nodal points occur where y(x, t) = 0. Since [tex]\( \cos(\omega t) \)[/tex] oscillates between -1 and 1, we need [tex]\( \sin(kx) = 0 \)[/tex] for the entire expression to be zero.
For [tex]\( \sin(kx) = 0 \)[/tex], kx must be a multiple of [tex]\( \pi \)[/tex] (since[tex]\( \sin(\pi n) = 0 \)[/tex] for integer n ). So, we have:
[tex]\[ kx = n \pi \][/tex]
Solving for x, we get:
[tex]\[ x = \frac{n \pi}{k} \][/tex]
Since [tex]\( k = \frac{2\pi}{\lambda} \)[/tex], where [tex]\( \lambda \)[/tex] is the wavelength, we can rewrite x as:
[tex]\[ x = \frac{n \lambda}{2} \][/tex]
Where n is an integer.
The first three nonzero nodal points occur when n = 1, 2, and 3. Substituting these values into the equation for x, we get:
1. [tex]\( x_1 = \frac{\lambda}{2} \)[/tex]
2. [tex]\( x_2 = \lambda \)[/tex]
3. [tex]\( x_3 = \frac{3\lambda}{2} \)[/tex]
So, the factors that multiply [tex]\( \lambda \)[/tex] in increasing order are [tex]\( \frac{1}{2}, 1, \) and \( \frac{3}{2} \)[/tex].
Let's summarize the steps:
1. Express displacement: Write down the displacement expression for a standing wave on a string.
2. Identify nodal points: Find where the displacement is zero by setting [tex]\( \sin(kx) = 0 \)[/tex].
3. Express nodal points: Use [tex]\( k = \frac{2\pi}{\lambda} \)[/tex] to express the nodal points in terms of the wavelength [tex]\( \lambda \)[/tex].
4. Determine factors of [tex]\( \lambda \)[/tex]: Identify the factors that multiply [tex]\( \lambda \)[/tex] in the nodal points.
5. List the factors: Arrange the factors in increasing order and list them.
The first three nonzero nodal points as multiples of the wavelength [tex]\( \lambda \)[/tex] are
[tex]\[ x_1 = \frac{1}{2} \lambda \][/tex]
[tex]\[ x_2 = \frac{2}{2} \lambda = \lambda \][/tex]
[tex]\[ x_3 = \frac{3}{2} \lambda \][/tex]
To find the first three nodal points closest to [tex]\(x = 0\)[/tex] with [tex]\(x > 0\)[/tex], where the displacement of the string [tex]\(y(x, t)\)[/tex] is zero for all times, we can use the formula for the nth nodal point of a standing wave on a string fixed at both ends:
[tex]\[ x_n = \frac{n}{2} \lambda \][/tex]
where \( n \) is the nodal number ([tex]1, 2, 3[/tex], ...), [tex]\( \lambda \)[/tex] is the wavelength of the wave, and [tex]\( x_n \)[/tex] is the nth nodal point.
Since we are looking for the first three nonzero nodal points, we will calculate [tex]\( x_1 \), \( x_2 \)[/tex], and [tex]\( x_3 \)[/tex] using [tex]\( n = 1 \), \( n = 2 \),[/tex] and [tex]\( n = 3 \),[/tex] respectively.
The factors that multiply [tex]\( \lambda \)[/tex] in increasing order for the first three nonzero nodal points are [tex]1[/tex], [tex]2[/tex], and [tex]3[/tex].
Therefore, the first three nonzero nodal points as multiples of the wavelength [tex]\( \lambda \)[/tex] are:
[tex]\[ x_1 = \frac{1}{2} \lambda \][/tex]
[tex]\[ x_2 = \frac{2}{2} \lambda = \lambda \][/tex]
[tex]\[ x_3 = \frac{3}{2} \lambda \][/tex]
These nodal points are located at [tex]\( \frac{1}{2} \)[/tex], [tex]1[/tex], and [tex]\( \frac{3}{2} \)[/tex] times the wavelength away from the origin, respectively.
The number of accidents per week at a hazardous intersection varies with mean 2.0 and standard deviation 1.2. This distribution takes only whole-number values, so it is certainly not normal. (a) Let x be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate distribution of x according to the Central Limit Theorem? μx = σx = (b) What is the approximate probability that x is less than 2? (c) What is the approximate probability that there are fewer than 120 accidents at the intersection in a year? (Hint: Restate this event in terms of x.)
Answer:
a.
μ= 2.0
σ/√n= 1.664
b.
P(X < 2) = 0.05
c.
P(X < 120)= 1
Step-by-step explanation:
Hello!
The study variable is:
X: Number of accidents that occur in an intersection during a period of 52 weeks.
This variable has an unknown distribution but it is known that is mean is μ= 2.0 and its standard deviation is σ= 1.2.
The central limit theorem states that if your sample is big enough (n ≥ 30)even if the distribution of the study variable is unknown, you can aproximate the distribution of the sample mean to normal, symbolically:
X[bar]≈N(μ; σ²/n)
The mean is the same value as the variable
μ= 2.0
And it's variance
σ²/n= (1.2)²/52= 2.769
Standard deviation
σ/√n= √(σ²/n)= 1.664
b.
P(X < 2) = P(Z < [(2 -2)/1.66])= P(Z< 0) = 0.5
c.
P(X < 120) = P(Z < [(120-2)/1.66]) = P(Z < 71.08) ≅ 1
I hope it helps!
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.