Question

a survey found that the american family generates an average of 17.2 pounds of glass garbage each year. assume the standard deviation of the distribution is 2.5 pounds. find the probability that the mean of a sample of 39 families will be between 17 and 18 pounds. assume that the sample is taken from a large population and the correction factor can be ignored. round your final answer to four decimal places and intermediate z - value calculations to two decimal places. $p(17<\overline{x}<18)=\square$

Explanation:

Step1: Calculate the standard error

The standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \), where \( \sigma = 2.5 \) and \( n = 39 \).
\( \sigma_{\bar{x}}=\frac{2.5}{\sqrt{39}}\approx\frac{2.5}{6.24}\approx0.40 \)

Step2: Calculate the z - scores

The z - score formula is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \)
For \( \bar{x}=17 \), \( z_1=\frac{17 - 17.2}{0.40}=\frac{- 0.2}{0.40}=- 0.50 \)
For \( \bar{x}=18 \), \( z_2=\frac{18 - 17.2}{0.40}=\frac{0.8}{0.40}=2.00 \)

Step3: Find the probabilities

Using the standard normal distribution table, \( P(Z < - 0.50)=0.3085 \), \( P(Z < 2.00)=0.9772 \)

Step4: Calculate the required probability

\( P(17<\bar{X}<18)=P(Z < 2.00)-P(Z < - 0.50) \)
\( P(17<\bar{X}<18)=0.9772 - 0.3085=0.6687 \)

Answer:

\( 0.6687 \)