Questions: A population of values has a normal distribution with μ=124.1 and σ=95.8. If a sample size of n=250 is selected, find the probability that the mean of the group of 250 is less than 141.1. Round to 3 decimal places. P(X<141.1)=

A population of values has a normal distribution with μ=124.1 and σ=95.8.

If a sample size of n=250 is selected, find the probability that the mean of the group of 250 is less than 141.1. Round to 3 decimal places.
P(X<141.1)=

Transcript text: A population of values has a normal distribution with $\mu=124.1$ and $\sigma=95.8$. If a sample size of $n=250$ is selected, find the probability that the mean of the group of 250 is less than 141.1. Round to 3 decimal places. \[ P(X<141.1)= \]

Solution

Solution Steps

Step 1: Define the Problem

We are given a normally distributed population with mean $ \mu = 124.1 $ and standard deviation $ \sigma = 95.8 $. We need to find the probability that the sample mean of a sample size $ n = 250 $ is less than $ 141.1 $.

Step 2: Calculate the Standard Error

The standard error of the mean (SEM) is calculated using the formula: \[ \sigma_{X} = \frac{\sigma}{\sqrt{n}} = \frac{95.8}{\sqrt{250}} \approx 6.058 \]

Step 3: Calculate the Z-scores

Next, we calculate the Z-scores for the sample mean:

For the upper bound $ X = 141.1 $: \[ Z_{end} = \frac{X - \mu}{\sigma_{X}} = \frac{141.1 - 124.1}{6.058} \approx 2.806 \]
For the lower bound, since we are considering values less than $ 141.1 $, we have: \[ Z_{start} = -\infty \]

Step 4: Calculate the Probability

Using the Z-scores, we can find the probability: \[ P(X < 141.1) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(2.806) - \Phi(-\infty) \approx 0.997 \]