Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 62 customers, answer the following questions.

(a)
What is the likelihood the sample mean is at least $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability

(b)
What is the likelihood the sample mean is greater than $25.00 but less than $27.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability

(c)
Within what limits will 90 percent of the sample means occur? (Round your answers to 2 decimal places.)

Answer 1

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

`\mu = 26, \sigma = 6, n = 62, s = (6)/(√(62)) = 0.762`

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (27 - 26)/(0.762)`

`Z = 1.31`

`Z = 1.31` has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

`Z = (X - \mu)/(s)`

`Z = (27 - 26)/(0.762)`

`Z = 1.31`

`Z = 1.31` has a pvalue of 0.9049

X = 25

`Z = (X - \mu)/(s)`

`Z = (25 - 26)/(0.762)`

`Z = -1.31`

`Z = -1.31` has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

`Z = (X - \mu)/(s)`

`-1.645 = (X - 26)/(0.762)`

`X - 26 = -1.645*0.762`

`X = 24.75`

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

`Z = (X - \mu)/(s)`

`1.645 = (X - 26)/(0.762)`

`X - 26 = 1.645*0.762`

`X = 27.25`

Between $24.75 and $27.25.