Question

In the country of United States of Heightlandia, the height measurements of ten-year-old children areapproximately normally distributed with a mean of 55.7 inches, and standard deviation of 1.9 inches.What is the probability that the height of a randomly chosen child is between 54.25 and 55.95 inches? Do notround until you get your your final answer, and then round to 3 decimal places.Answers(Round your answer to 3 decimal places.)

Answer 1

Firstly, we will calculate the z-score for each value => 54.25 and 55.95 inches.

The formula to calculate the z-score will be

`Z=(x-\mu)/(\sigma)`
`\begin{gathered} \text{where} \\ x=\text{ sample value} \\ \mu=\text{ mean} \\ \sigma=\text{ Standard deviation} \end{gathered}`

For x = 54.25, the z-score will be

`\begin{gathered} Z_1=(54.25-55.7)/(1.9) \\ =-(29)/(38)\approx-0.763 \end{gathered}`

For x = 55.95, the z-score will be

`\begin{gathered} Z_2=(55.95-55.7)/(1.9) \\ =0.132 \end{gathered}`

The probability is the shaded portion of the image above.

Therefore,

`\begin{gathered} P(Z_1\leq Z\leq Z_2)=P(-0.763\leq Z\leq0.132) \\ \text{Hence,} \\ P=P(0\leq Z\leq0.132)+P(0\leq Z\leq0.763) \\ =0.0525+0.2773 \\ =0.3298 \\ P\approx0.330\text{ in 3d.p} \end{gathered}`

Therefore, the probability is 0.330 in 3 d.p.