Question

x is a normally distributed random variable with mean 64 and standard deviation 14. what is the probability that x is between 63 and 95? 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 \(\mu = 64\) (mean) and \(\sigma=14\) (standard deviation).
For \(x = 63\):
\(z_1=\frac{63 - 64}{14}=\frac{-1}{14}\approx - 0.071\)
For \(x = 95\):
\(z_2=\frac{95 - 64}{14}=\frac{31}{14}\approx2.214\)

Step2: Use the standard normal distribution table

We want to find \(P(-0.071<Z<2.214)\).
Since \(P(-0.071 < Z<2.214)=P(Z < 2.214)-P(Z<-0.071)\)
From the standard - normal table: \(P(Z < 2.214)\approx0.986\), \(P(Z<-0.071)\approx0.472\)

Step3: Calculate the probability

\(P(-0.071 < Z<2.214)=0.986 - 0.472=0.514\)

Answer:

\(0.514\)