Question

x is a normally distributed random variable with mean 60 and standard deviation 12. what is the probability that x is between 32 and 88? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 60\), \(\sigma=12\).
For \(x = 32\):
\(z_1=\frac{32 - 60}{12}=\frac{-28}{12}\approx - 2.33\)
For \(x = 88\):
\(z_2=\frac{88 - 60}{12}=\frac{28}{12}\approx2.33\)

Step2: Find the probabilities using the standard normal distribution table

We know that \(P(32<X<88)=P(-2.33<Z<2.33)\)
Since \(P(-2.33<Z<2.33)=\Phi(2.33)-\Phi(-2.33)\)
From the standard normal table, \(\Phi(2.33)=0.9901\) and \(\Phi(-2.33)=1 - \Phi(2.33)=1 - 0.9901 = 0.0099\)
\(P(-2.33<Z<2.33)=0.9901-0.0099 = 0.9802\)

Answer:

\(0.980\)