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 56.3 inches, and standard deviation of 4.9 inches.
what is the probability that the height of a randomly chosen child is between 55.25 and 63.45 inches?
use the normal table and give answer to 4 decimal places
answer=

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 56.3\), \(\sigma=4.9\).
For \(x = 55.25\):
\(z_1=\frac{55.25 - 56.3}{4.9}=\frac{-1.05}{4.9}\approx - 0.21\)
For \(x = 63.45\):
\(z_2=\frac{63.45 - 56.3}{4.9}=\frac{7.15}{4.9}\approx1.46\)

Step2: Use the standard normal table

We want to find \(P(-0.21<Z<1.46)\).
We know that \(P(-0.21 < Z<1.46)=P(Z < 1.46)-P(Z<-0.21)\)
From the standard normal table: \(P(Z < 1.46)=0.9279\), \(P(Z<-0.21) = 0.4168\)

Step3: Calculate the probability

\(P(-0.21<Z<1.46)=0.9279 - 0.4168=0.5111\)

Answer:

\(0.5111\)