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 54.6 inches, and standard deviation of 1.4 inches.

What is the probability that the height of a randomly chosen child is between 52.9 and 55 inches? Do not round until you get your your final answer, and then round to 3 decimal places.

Answer 1

Let the variable X denotes height measurements of ten year old children.

X follows Normal distribution with mean = 54.6 inches and standard deviation = 1.4 inches.

We have to find probability that height of randomly chosen child is between 52.9 and 55 inches.

That is P(52.9 < X < 55).

By using z score formula, $Z = \frac{X - \mu }{\sigma }$

Where, $\mu$ is mean and $\sigma$ is standard deviation.

P(52.9 < X < 55) = $P(\frac{52.9-54.6}{1.4}< \frac{X - \mu }{\sigma } < \frac{55-54.6}{1.4})$

= P(-1.2143 < Z < 0.2857)

= P( Z < 0.2857) - P(Z < -1.2143)

= 0.612445 - 0.112316 .................... (Using Excel)

= 0.500129

~ 0.500

Hence, P(52.9 < X < 55) = 0.500