Question

the length of zebra pregnancies is normally distributed, with mean (mu = 380) and standard deviation (sigma = 10). a random sample of 20 random pregnant zebras is chosen. find (p(overline{x}<375)) for (n = 20). enter your answer as an area under the curve with 4 decimal places. (p(overline{x}<375)=)

Explanation:

Step1: Calculate the standard error

The standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \), where \( \sigma = 10 \) and \( n = 20 \).
\( \sigma_{\bar{x}}=\frac{10}{\sqrt{20}}\approx2.2361 \)

Step2: Calculate the z - score

The z - score formula is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \), where \( \bar{x}=375 \), \( \mu = 380 \) and \( \sigma_{\bar{x}}\approx2.2361 \)
\( z=\frac{375 - 380}{2.2361}=\frac{- 5}{2.2361}\approx - 2.24 \)

Step3: Find the probability

Using the standard normal distribution table or a calculator with a normal - distribution function (e.g., in R: pnorm(-2.24) or in Excel: NORM.S.DIST(-2.24,TRUE)), we find \( P(Z\lt - 2.24) \)

Answer:

\(0.0125\)