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.

Question

asked Dec 18, 2019 193k views

2 Answers

Jcady · Answer 1 · 2019-12-18T21:58:25+0000

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

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.

$Z=(X-\mu)/(\sigma),$

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.

$-0.25=(57-\mu)/(\sigma)\\\\-0.25\sigma=57-\mu---------(1)$

$1.25=(87-\mu)/(\sigma)\\\\1.25\sigma=87-\mu----------(2)$

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

$1.50\sigma=30\\\\\Rightarrow\ \sigma=(30)/(1.5)=20$

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.

Chris Jacob · Answer 2 · 2019-12-22T21:13:02+0000

Here is dependence between scores and x-values:

$Z_i=(X_i-\mu)/(\sigma),$

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

$-0.25=(57-\mu)/(\sigma).$

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

$1.25=(87-\mu)/(\sigma).$

Now solve the system of equations:

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

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

Subtract first equation from the second:

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

Then

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

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