Question

in the country of united states of heightlandia, the height measurements of ten - year - old children are approximately normally distributed with a mean of 53.1 inches, and standard deviation of 8.3 inches.
what is the probability that the height of a randomly chosen child is between 52.65 and 61.35 inches? do not round until you get your your final answer, and then round to 3 decimal places.
answer=
(round your answer to 3 decimal places.)
question help: video

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 53.1\), \(\sigma=8.3\)

For \(x = 52.65\):
\(z_1=\frac{52.65 - 53.1}{8.3}=\frac{-0.45}{8.3}\approx - 0.054\)

For \(x = 61.35\):
\(z_2=\frac{61.35 - 53.1}{8.3}=\frac{8.25}{8.3}\approx0.994\)

Step2: Use the standard normal distribution table

We want to find \(P(-0.054<Z<0.994)\)

Since \(P(-0.054 < Z<0.994)=P(Z < 0.994)-P(Z<-0.054)\)

From the standard normal table, \(P(Z < 0.994)\approx0.839\), \(P(Z<-0.054)\approx0.478\)

\(P(-0.054 < Z<0.994)=0.839 - 0.478=0.361\)

Answer:

\(0.361\)