Question

problem 6
a normal population has a mean $\mu = 43$ and standard deviation is 9. what is the probability that a randomly chosen value will be greater than 25?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(x = 25\), \(\mu=43\), and \(\sigma = 9\).

$$z=\frac{25 - 43}{9}=\frac{- 18}{9}=-2$$

Step2: Find the probability

We want \(P(X>25)\). Since \(P(X>25)=1 - P(X\leq25)\), and for a normal distribution \(P(X\leq x)=\Phi(z)\) (where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution).
From the standard normal table, \(\Phi(-2)=0.0228\)
So \(P(X>25)=1 - 0.0228 = 0.9772\)

Answer:

\(0.9772\)