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

Answer:

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762$

(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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ 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

$Z = \frac{X - \mu}{s}$

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ has a pvalue of 0.9049

X = 25

$Z = \frac{X - \mu}{s}$

$Z = \frac{25 - 26}{0.762}$

$Z = -1.31$

$Z = -1.31$ 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

$Z = \frac{X - \mu}{s}$

$-1.645 = \frac{X - 26}{0.762}$

$X - 26 = -1.645*0.762$

$X = 24.75$

95th percentile

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

$Z = \frac{X - \mu}{s}$

$1.645 = \frac{X - 26}{0.762}$

$X - 26 = 1.645*0.762$

$X = 27.25$

Between $24.75 and $27.25.