Question

1. (4 points) consider a population with the mean \\( \mu = 51,000 \\) and the standard deviation \\( \sigma = 5,000 \\). for a random sample of size \\( n = 100 \\), find the probability that the sample mean \\( x \\) will be between 50,000 and 51,500.

Explanation:

Step1: Calculate the standard error

The standard error $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 5000$ and $n = 100$, then $\sigma_{\bar{X}}=\frac{5000}{\sqrt{100}}=\frac{5000}{10}=500$.

Step2: Standardize the sample - mean values

For $x_1 = 50000$, the z - score $z_1=\frac{\bar{x}_1-\mu}{\sigma_{\bar{X}}}=\frac{50000 - 51000}{500}=\frac{-1000}{500}=-2$.
For $x_2 = 51500$, the z - score $z_2=\frac{\bar{x}_2-\mu}{\sigma_{\bar{X}}}=\frac{51500 - 51000}{500}=\frac{500}{500}=1$.

Step3: Find the probability

We want to find $P(50000<\bar{X}<51500)$, which is equivalent to $P(-2 < Z<1)$.
Using the standard normal distribution property $P(-2 < Z<1)=\Phi(1)-\Phi(-2)$.
From the standard normal table, $\Phi(1)=0.8413$ and $\Phi(-2)=0.0228$.
So $P(-2 < Z<1)=0.8413-0.0228 = 0.8185$.

Answer:

The probability that the sample mean $\bar{X}$ will be between $50000$ and $51500$ is $0.8185$.