Question

Out of 2,000 random but normally distributed numbers with a mean of 45 and a standard deviation of x, approximately 1,360 numbers are found to be between 40 and 50. What is the value of x?

Answer 1

Let $Y$ denote the random variable representing a given number in the total set of numbers. We're told that $\dfrac{1360}{2000}=0.68$ of the numbers fall within a given range, so we know

$\mathbb P(40\le Y\le50)=0.68$

where $Y$ is normally distributed with mean 45 and an unknown variance $x^2$.

Let's make the transformation to a random variable with a standard normal distribution:

$\mathbb P\left(\dfrac{40-45}x\le\dfrac{Y-45}x\le\dfrac{50-45}x\right)=\mathbb P\left(-\dfrac5x\le Z\le\dfrac5x\right)$

Since $Z$ is symmetric, we have

$\mathbb P(-\dfrac5x\le Z\le\dfrac5x\right)=2\mathbb P\left(0\le Z\le\dfrac5x\right)=2\bigg(\mathbb P\left(Z\le\dfrac5x\right)-\mathbb P(Z\le0)\bigg)$

The mean of $Z$ is 0, and by symmetry we know that exactly half of the distribution falls to the left of $Z=0$, so $\mathbb P(Z\le0)=0.5$. We're left with

$0.68=2\mathbb P\left(Z\le\dfrac5x\right)-2\cdot0.5$
$\implies\mathbb P\left(Z\le\dfrac5x\right)=0.84$

This probability corresponds to a value of $Z\approx0.9945$, which means

$\dfrac5x\approx0.9945\implies x\approx5.0277$