Question

x is a normally distributed random variable with mean 60 and standard deviation 18. what is the probability that x is between 47 and 73? write your answer as a decimal rounded to the nearest thousandth. save answer

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 60\) (mean) and \(\sigma=18\) (standard deviation).
For \(x = 47\):
\(z_1=\frac{47 - 60}{18}=\frac{-13}{18}\approx - 0.72\)
For \(x = 73\):
\(z_2=\frac{73 - 60}{18}=\frac{13}{18}\approx0.72\)

Step2: Use the standard normal distribution table

We want to find \(P(-0.72<Z<0.72)\).
Since the standard normal distribution is symmetric, \(P(-0.72 < Z<0.72)=2\Phi(0.72)-1\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.
From the standard - normal table, \(\Phi(0.72)\approx0.7642\)
Then \(2\times0.7642-1=1.5284 - 1=0.5284\approx0.528\)

Answer:

\(0.528\)