Question

The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 50 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.3 pounds, what is the probability that the sample mean will be each of the following? a. More than 59 pounds b. More than 56 pounds c. Between 56 and 57 pounds d. Less than 53 pounds e. Less than 49 pounds *Round the values of z to 2 decimal places. Round your answer to 4 decimal places, the tolerance is +/-0.0001. **Round the values of z to 2 decimal places. Round your answer to 4 decimal places. a. * b. * c. * d. * e. **

Answer 1

Answer:

a. 0.1038 = 10.38% probability that the sample mean is more than 59 pounds.

b. 0.6772 = 67.72% probability that the sample mean is more than 56 pounds.

c. 0.2210 = 22.10% probability that the sample mean is between 56 and 57 pounds.

d. 0.0146 = 1.46% probability that the sample mean is less than 53 pounds.

e. 0% probability that the sample mean is less than 49 pounds.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, 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.

The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. The population standard deviation of annual usage is 12.3 pounds.

This means that $\mu = 56.8, \sigma = 12.3$

Sample of 50:

This means that $n = 50, s = \frac{12.3}{\sqrt{50}} = 1.74$

a. More than 59 pounds

This is 1 subtracted by the pvalue of Z when X = 59.

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

By the Central Limit Theorem

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

$Z = \frac{59 - 56.8}{1.74}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962

1 - 0.8962 = 0.1038

0.1038 = 10.38% probability that the sample mean is more than 59 pounds.

b. More than 56 pounds

This is 1 subtracted by the pvalue of Z when X = 56. So

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

$Z = \frac{56 - 56.8}{1.74}$

$Z = -0.46$

$Z = -0.46$ has a pvalue of 0.3228

1 - 0.3228 = 0.6772

0.6772 = 67.72% probability that the sample mean is more than 56 pounds.

c. Between 56 and 57 pounds

This is the pvalue of Z when X = 57 subtracted by the pvalue of Z when X = 56.

X = 57

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

$Z = \frac{57 - 56.8}{1.74}$

$Z = 0.11$

$Z = 0.11$ has a pvalue of 0.5438

X = 56

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

$Z = \frac{56 - 56.8}{1.74}$

$Z = -0.46$

$Z = -0.46$ has a pvalue of 0.3228

0.5438 - 0.3228 = 0.2210

0.2210 = 22.10% probability that the sample mean is between 56 and 57 pounds.

d. Less than 53 pounds

This is the pvalue of Z when X = 53.

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

$Z = \frac{53 - 56.8}{1.74}$

$Z = -2.18$

$Z = -2.18$ has a pvalue of 0.0146

0.0146 = 1.46% probability that the sample mean is less than 53 pounds.

e. Less than 49 pounds

This is the pvalue of Z when X = 49.

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

$Z = \frac{49 - 56.8}{1.74}$

$Z = -4.48$

$Z = -4.48$ has a pvalue of 0

0% probability that the sample mean is less than 49 pounds.