Question

in an all boys school, the heights of the student body are normally distributed with a mean of 68 inches and a standard deviation of 2.5 inches. what percentage of the students are between 65 and 72 inches tall, to the nearest tenth?
statistics calculator

Explanation:

Step1: Identify the mean and standard deviation

The mean ($\mu$) is 68 inches and the standard deviation ($\sigma$) is 2.5 inches. We need to find the percentage of students with heights between 65 and 72 inches.

Step2: Calculate the z - scores

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$.

For $x = 65$:
$z_1=\frac{65 - 68}{2.5}=\frac{- 3}{2.5}=- 1.2$

For $x = 72$:
$z_2=\frac{72 - 68}{2.5}=\frac{4}{2.5} = 1.6$

Step3: Find the area between the z - scores

We use the standard normal distribution table (or a calculator with normal distribution functions).

The area to the left of $z=-1.2$ is $P(Z < - 1.2)$. Looking up in the standard normal table, $P(Z < - 1.2)=0.1151$.

The area to the left of $z = 1.6$ is $P(Z < 1.6)$. Looking up in the standard normal table, $P(Z < 1.6)=0.9452$.

The area between $z=-1.2$ and $z = 1.6$ is $P(-1.2

$P(-1.2

Answer:

The percentage of students between 65 and 72 inches tall is approximately $83.0\%$ (to the nearest tenth).