Question

An economist uses the price of a gallon of milk as a measure of inflation. She finds that the average price is $3.82 per gallon and the population standard deviation is $0.33. You decide to sample 40 convenience stores, collect their prices for a gallon of milk, and compute the mean price for the sample.
a. what is the standard error of the mean in this experiment?

b. what is the probability that the sample mean is between $3.78 and $3.86?

c. what is the probability that the difference between the sample mean and the population mean is less than $0.01?

d. what is the likelihood the sample mean is greater than $3.92?

Answer 1

Answer:

(a) The standard error of the mean in this experiment is $0.052.

(b) The probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c) The probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d) The likelihood that the sample mean is greater than $3.92 is 0.9726.

Step-by-step explanation:

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the distribution of sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The information provided is:

$n=40\\\mu=\ 3.82\\\sigma=\ 0.33$

As n = 40 > 30, the distribution of sample mean is $\bar X\sim N(3.82,\ 0.052^{2})$.

(a)

The standard error is the standard deviation of the sampling distribution of sample mean.

Compute the standard deviation of the sampling distribution of sample mean as follows:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

    $=\frac{0.33}{\sqrt{40}}\\\\=0.052178\\\\\approx 0.052$

Thus, the standard error of the mean in this experiment is $0.052.

(b)

Compute the probability that the sample mean is between $3.78 and $3.86 as follows:

$P(3.78$

                               $=P(-0.77$

Thus, the probability that the sample mean is between $3.78 and $3.86 is 0.5587.

(c)

If the difference between the sample mean and the population mean is less than $0.01 then:

$\bar X-\mu_{\bar x}$

Compute the value of $P(\bar X$ as follows:

$P(\bar X$

                    $=P(Z$

Thus, the probability that the difference between the sample mean and the population mean is less than $0.01 is 0.5754.

(d)

Compute the probability that the sample mean is greater than $3.92 as follows:

$P(\bar X>3.92)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{3.92-3.82}{0.052})$

                    $=P(Z$

Thus, the likelihood that the sample mean is greater than $3.92 is 0.9726.