Question

In a population distribution, a score of x=57 corresponds to z=-0.25 and a score of x=87 corresponds to z=1.25. Find the mean and standard deviation for the population.

Answer 1

Answer: The mean is 62 and the standard deviation is 20 for the population.

Step-by-step explanation:

Let $\mu$ be the mean and $\sigma$ be the standard deviation .

Formula to calculate z-score corresponds to random variable x on normal curve.

$Z=\dfrac{X-\mu}{\sigma},$  

Given : In a population distribution, a score of x=57 corresponds to z=-0.25 and a score of x=87 corresponds to z=1.25.

$-0.25=\dfrac{57-\mu}{\sigma}\\\\-0.25\sigma=57-\mu---------(1)$

$1.25=\dfrac{87-\mu}{\sigma}\\\\1.25\sigma=87-\mu----------(2)$

Eliminate equation(1) from equation(2), we get

$1.50\sigma=30\\\\\Rightarrow\ \sigma=\dfrac{30}{1.5}=20$

Put value of $\sigma=20$ in (1)

$1.25(20)=87-\mu\\\\87-\mu=25,\\\\\Rightarrow\mu=87-25=62.$

Hence, the mean is 62 and the standard deviation is 20 for the population.

Answer 2

Here is dependence between scores and x-values:

$Z_i=\dfrac{X_i-\mu}{\sigma},$

where $\mu$ is the mean, $\sigma$ is standard deviation and i changes from 1 to 2.

1. When i=1, $Z_1=-0.25,\ X_1=57,$ then

$-0.25=\dfrac{57-\mu}{\sigma}.$

2. When i=2, $Z_2=1.25,\ X_2=87,$ then

$1.25=\dfrac{87-\mu}{\sigma}.$

Now solve the system of equations:

$\left\{\begin{array}{l}-0.25=\dfrac{57-\mu}{\sigma}\\ \\1.25=\dfrac{87-\mu}{\sigma}.\end{array}\right.$

$\left\{\begin{array}{l}-0.25\sigma=57-\mu\\ \\1.25\sigma=87-\mu.\end{array}\right.$

Subtract first equation from the second:

$1.25\sigma-(-0.25\sigma)=87-57,\\ \\1.5\sigma=30,\\ \\\sigma=20.$

Then

$1.25=\dfrac{87-\mu}{20},\\ \\87-\mu=25,\\ \\\mu=87-25=62.$

Answer: the mean is 62, the standard deviation is 20.