Question

a variable is normally distributed with mean 7 and standard deviation 2.
a. find the percentage of all possible values of the variable that lie between 5 and 11.
b. find the percentage of all possible values of the variable that exceed 4.
c. find the percentage of all possible values of the variable that are less than 3.

Explanation:

Step1: Standardize the values

For a normal distribution \(X\sim N(\mu = 7,\sigma = 2)\), the z - score is calculated as \(z=\frac{x-\mu}{\sigma}\)

Part a:

For \(x_1 = 5\), \(z_1=\frac{5 - 7}{2}=\frac{-2}{2}=-1\)
For \(x_2 = 11\), \(z_2=\frac{11 - 7}{2}=\frac{4}{2} = 2\)
We want to find \(P(5<X<11)=P(-1<Z<2)\)
Using the property \(P(-1<Z<2)=P(Z < 2)-P(Z<-1)\)
From the standard normal table, \(P(Z < 2)=0.9772\), \(P(Z<-1)=0.1587\)
\(P(-1 < Z < 2)=0.9772-0.1587 = 0.8185\)

Part b:

For \(x = 4\), \(z=\frac{4 - 7}{2}=\frac{-3}{2}=-1.5\)
We want to find \(P(X>4)=1 - P(X\leq4)\)
\(P(X\leq4)\) corresponds to \(P(Z\leq-1.5)\)
From the standard - normal table, \(P(Z\leq-1.5)=0.0668\)
\(P(X > 4)=1-0.0668 = 0.9332\)

Part c:

For \(x = 3\), \(z=\frac{3 - 7}{2}=\frac{-4}{2}=-2\)
We want to find \(P(X < 3)\)
From the standard - normal table, \(P(Z<-2)=0.0228\)

Answer:

a. \(81.85\%\)
b. \(93.32\%\)
c. \(2.28\%\)