Question

question 21 of 25
suppose a normal distribution has a mean of 62 and a standard deviation of

1. what is the probability that a data value is between 57 and 65? round your

answer to the nearest tenth of a percent.

a. 67.8%
b. 69.8%
c. 68.8%
d. 66.8%

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 62\), \(\sigma=4\).
For \(x = 57\):
\(z_1=\frac{57 - 62}{4}=\frac{-5}{4}=-1.25\)
For \(x = 65\):
\(z_2=\frac{65 - 62}{4}=\frac{3}{4}=0.75\)

Step2: Use the standard normal distribution table

We know that \(P(57<X<65)=P(-1.25<Z<0.75)\)
Since \(P(-1.25 < Z<0.75)=P(Z < 0.75)-P(Z<-1.25)\)
From the standard normal distribution table: \(P(Z < 0.75)=0.7734\), \(P(Z<-1.25)=0.1056\)

Step3: Calculate the probability

\(P(-1.25 < Z<0.75)=0.7734 - 0.1056=0.6678\)

Step4: Convert to percentage

Multiply by \(100\) to get the percentage: \(0.6678\times100 = 66.78\%\approx66.8\%\)

Answer:

D. 66.8%