Question

the length of zebra pregnancies is normally distributed, with mean \\( \mu = 380 \\) and standard deviation \\( \sigma = 10 \\). a random sample of 11 random pregnant zebras is chosen. find \\( p( \bar{x} < 375) \\) for \\( n = 11 \\). enter your answer as an area under the curve with 4 decimal places. \\( p( \bar{x} < 375)= \\)

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 10$ and $n = 11$, we have $\sigma_{\bar{x}}=\frac{10}{\sqrt{11}}\approx3.0151$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
Here, $\bar{x}=375$, $\mu = 380$, and $\sigma_{\bar{x}}\approx3.0151$.
So, $z=\frac{375 - 380}{3.0151}=\frac{-5}{3.0151}\approx - 1.66$.

Step3: Find the probability

We want to find $P(\bar{X}<375)$, which is equivalent to $P(Z < - 1.66)$ using the standard normal distribution.
Looking up the value in the standard normal table (or using a calculator with a normal - distribution function), we find that $P(Z < - 1.66)=0.0485$.

Answer:

$0.0485$