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 55.7 inches, and standard deviation of 5.2 inches. What is the probability that the height of a randomly chosen child is between 49.5 and 67.2 inches? Do not round until you get your your final answer, and then round to 3 decimal places.

Answer 1

Answer:

$P(49.5<X<67.2)=P(\frac{49.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{67.2-\mu}{\sigma})=P(\frac{49.5-55.7}{5.2}<Z<\frac{67.2-55.7}{5.2})=P(-1.192<z<2.212)$

$P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)$

$P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)=0.987-0.117=0.870$

Step-by-step explanation:

We define X the random variable that represent the heights of a population for ten year old children, and for this case we know the distribution for X is given by:

$X \sim N(55.7,5.2)$  

Where $\mu=55.7$ and $\sigma=5.2$

We want to find this probability:

$P(49.5<X<67.2)$

We can use the z score formula to solve this problem given by:

$z=\frac{x-\mu}{\sigma}$

Using this formula we have:

$P(49.5<X<67.2)=P(\frac{49.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{67.2-\mu}{\sigma})=P(\frac{49.5-55.7}{5.2}<Z<\frac{67.2-55.7}{5.2})=P(-1.192<z<2.212)$

And we can find this probability with this difference

$P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)$

We can use tables for the normal standard distribution, excel or a calculator and we got this

$P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)=0.987-0.117=0.870$