Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is $27.50 with a standard deviation of $7.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 68 customers, answer the following questions

a. What is the likelihood the sample mean is at least $30.00?
b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?
c. Within what limits will 90 percent of the sample means occur?

Answer 1

Answer:

a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.

b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00

c) 90% of sample means will occur between $26.1 and $28.9.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 question, we have that:

$\mu = 27.50, \sigma = 7, n = 68, s = \frac{7}{\sqrt{68}} = 0.85$

a. What is the likelihood the sample mean is at least $30.00?

This is 1 subtracted by the pvalue of Z when X = 30. So

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

By the Central Limit Theorem, we have that:

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

$Z = \frac{30 - 27.5}{0.85}$

$Z = 2.94$

$Z = 2.94$ has a pvalue of 0.9984

1 - 0.9984 = 0.0016

0.0016 = 0.16% probability that the sample mean is at least $30.00.

b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?

This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So

From a, when X = 30, Z has a pvalue of 0.9984

When X = 26.5

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

$Z = \frac{26.5 - 27.5}{0.85}$

$Z = -1.18$

$Z = -1.18$ has a pvalue of 0.1190

0.9984 - 0.1190 = 0.8794

0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.

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

Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645

Lower bound:

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

$-1.645 = \frac{X - 27.5}{0.85}$

$X - 27.5 = -1.645*0.85$

$X = 26.1$

Upper Bound:

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

$1.645 = \frac{X - 27.5}{0.85}$

$X - 27.5 = 1.645*0.85$

$X = 28.9$

90% of sample means will occur between $26.1 and $28.9.