Question

x is a normally distributed random variable with mean 83 and standard deviation 20. what is the probability that x is between 41 and 92? 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 \(\mu = 83\) (mean) and \(\sigma=20\) (standard deviation).
For \(x = 41\):
\(z_1=\frac{41 - 83}{20}=\frac{-42}{20}=-2.1\)
For \(x = 92\):
\(z_2=\frac{92 - 83}{20}=\frac{9}{20}=0.45\)

Step2: Use the standard normal distribution table

We want to find \(P(41<X<92)=P(-2.1<Z<0.45)\)
Since \(P(-2.1<Z<0.45)=P(Z < 0.45)-P(Z<-2.1)\)
From the standard normal table, \(P(Z < 0.45)=0.6736\) and \(P(Z<-2.1)=0.0179\)

Step3: Calculate the probability

\(P(-2.1<Z<0.45)=0.6736 - 0.0179=0.6557\)

Answer:

\(0.656\)