Question

x is a normally distributed random variable with mean 53 and standard deviation 17. what is the probability that x is between 13 and 93? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 53\), \(\sigma=17\).
For \(x = 13\):
\(z_1=\frac{13 - 53}{17}=\frac{-40}{17}\approx - 2.35\)
For \(x = 93\):
\(z_2=\frac{93 - 53}{17}=\frac{40}{17}\approx2.35\)

Step2: Find probabilities using the standard normal distribution table

We want to find \(P(-2.35<Z<2.35)\).
Since \(P(-2.35 < Z < 2.35)=P(Z < 2.35)-P(Z<-2.35)\)
From the standard normal distribution table, \(P(Z < 2.35)=0.9906\) and \(P(Z<-2.35) = 0.0094\)

Step3: Calculate the final probability

\(P(-2.35<Z<2.35)=0.9906 - 0.0094=0.9812\approx0.981\)

Answer:

\(0.981\)