Question

In: Statistics and Probability

The mean amount purchased by each customer at Summit bookstore is $35 with a standard deviation...

The mean amount purchased by each customer at Summit bookstore is $35 with a standard deviation of $9. The population is positively skewed. For a sample of 42 customers, answer the following questions: a. What is the likelihood the sample mean is at least $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean b. What is the likelihood the sample mean is greater than $33 but less than $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean c. Within what limits will 98% of the sample means occur? (Round the final answers to 2 decimal places.)

Solutions

Expert Solution

a.

The following information has been provided:

We need to compute Pr(X≥39). The corresponding z-value needed to be computed is:

Therefore, we get that

The following is obtained graphically:

b.

We need to compute . The corresponding z-values needed to be computed are:

Therefore, we get:

The following is obtained graphically:

c.

The answer is between:

31.76933

and

38.23067

Using excel:

=NORM.INV(0.02/2,35,1.38873014965883)

and

=NORM.INV(1-0.02/2,35,1.38873014965883)

The following is obtained graphically:

Please do upvote if you are satisfied! Let me know in the comments if anything is not clear. I will reply ASAP!


Related Solutions

The mean amount purchased by each customer at Churchill’s Grocery Store is $35 with a standard...
The mean amount purchased by each customer at Churchill’s Grocery Store is $35 with a standard deviation of $9. The population is positively skewed. For a sample of 42 customers, answer the following questions: a. What is the likelihood the sample mean is at least $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean             b. What is the likelihood the sample mean is greater than $33 but less than...
The mean amount purchased by each customer at Churchill’s Grocery Store is $28 with a standard...
The mean amount purchased by each customer at Churchill’s Grocery Store is $28 with a standard deviation of $7. The population is positively skewed. For a sample of 49 customers, answer the following questions: a. What is the likelihood the sample mean is at least $29? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean             b. What is the likelihood the sample mean is greater than $26 but less than...
The mean amount purchased by each customer at Churchill’s Grocery Store is $27 with a standard...
The mean amount purchased by each customer at Churchill’s Grocery Store is $27 with a standard deviation of $9. The population is positively skewed. For a sample of 48 customers, answer the following questions: a. What is the likelihood the sample mean is at least $29? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) b. What is the likelihood the sample mean is greater than $26 but less than $29? (Round the z-value...
The mean amount purchased by each customer at Churchill’s Grocery Store is $33 with a standard...
The mean amount purchased by each customer at Churchill’s Grocery Store is $33 with a standard deviation of $9. The population is positively skewed. For a sample of 56 customers, answer the following questions: a. What is the likelihood the sample mean is at least $34? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean: b. What is the likelihood the sample mean is greater than $30 but less than $34? (Round...
For a population with a mean equal to 250 and a standard deviation equal to 35​,...
For a population with a mean equal to 250 and a standard deviation equal to 35​, calculate the standard error of the mean for the following sample sizes. ​a) 10 ​b)  40 ​c)  70 The standard error of the mean for a sample size of 10 is The standard error of the mean for a sample size of 40 is The standard error of the mean for a sample size of 70 is
The mean and the standard deviation of the sample of 100 bank customer waiting times are...
The mean and the standard deviation of the sample of 100 bank customer waiting times are x⎯⎯x¯ = 5.33 and s = 2.008. Calculate a t-based 95 percent confidence interval for µ, the mean of all possible bank customer waiting times using the new system. Are we 95 percent confident that µ is less than 6 minutes?. (Choose the nearest degree of freedom for the given sample size. Round your answers to 3 decimal places.) The t-based 95 percent confidence...
Indicate the mean and the standard deviation of the distribution of means for each of the...
Indicate the mean and the standard deviation of the distribution of means for each of the following situations.     Population Sample Size Mean Variance N ​(a)80 80 5 (b) 80 15 5 ​(c) 80 10 5 (d) 80 50 25 ​(e) 80 5 25 Question 11 (1 point) In situation (a) above, the mean of the distribution of means is _____ and the standard deviation of the distribution of means is ______. In situation (b) above, the mean of the...
Let x be a normal random variable with mean 35 and standard deviation 2. a. Find...
Let x be a normal random variable with mean 35 and standard deviation 2. a. Find P(30 < x < 38). .9270 b. Find P(x > 34). .6915 c. Find P(x = 36). 0 d. Find the area under the distribution of x to the left of 31. .0228
Changing the mean and standard deviation of a Normal distribution by a moderate amount can greatly...
Changing the mean and standard deviation of a Normal distribution by a moderate amount can greatly change the percent of observations in the tails. Suppose that a college is looking for applicants with SAT math scores 750 and above. (a) In 2015, the scores of men on the math SAT followed the N ( 527 , 124 ) distribution. What percent of men scored 750 or better? A) 96.41 % B) 1.36 % C) 0.96 % D) 1.80 % E)...
The amount of time a bank teller spends with each customer has a population mean of...
The amount of time a bank teller spends with each customer has a population mean of 3.086 minutes. You select a random sample of 16 customers. The sample standard deviation is 0.40 minutes. There is a 95% chance that the sample mean is below ______ minutes
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT