Question

question 3 of 10
suppose a normal distribution has a mean of 26 and a standard deviation of

1. what is the probability that a data value is between 28 and 31? round your

answer to the nearest tenth of a percent.
a. 19.3%
b. 22.3%
c. 21.3%
d. 20.3%

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 26\), \(\sigma=4\).
For \(x = 28\):
\(z_1=\frac{28 - 26}{4}=\frac{2}{4}=0.5\)
For \(x = 31\):
\(z_2=\frac{31 - 26}{4}=\frac{5}{4}=1.25\)

Step2: Use the standard normal distribution table

We know that \(P(28\lt X\lt31)=P(0.5\lt Z\lt1.25)\)
\(P(0.5\lt Z\lt1.25)=P(Z\lt1.25)-P(Z\lt0.5)\)
From the standard normal distribution table, \(P(Z\lt0.5)=0.6915\) and \(P(Z\lt1.25)=0.8944\)

Step3: Calculate the probability

\(P(0.5\lt Z\lt1.25)=0.8944 - 0.6915=0.2029\)

Step4: Convert to percentage

\(0.2029\times100 = 20.29\%\approx20.3\%\)

Answer:

D. 20.3%