Question

problem 5
a normal population has a mean $\mu = 121$ and standard deviation is 25. what is the probability that a randomly chosen value will be less than 101?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 101\), \(\mu=121\), and \(\sigma = 25\).

$$z=\frac{101 - 121}{25}=\frac{-20}{25}=- 0.8$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt101)\), which is equivalent to \(P(Z\lt - 0.8)\) using the standard normal transformation.
Looking up the value of \(z=-0.8\) in the standard - normal distribution table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)\)), we find that \(P(Z\lt - 0.8)=0.2119\)

Answer:

\(0.2119\)