Answer to Question #212300 in Economics of Enterprise for Ramya

Question

In a normal distribution 10% of the items are under 50 and 90% of the items are under 

80. Find the values of mean and standard deviation of the distribution.

Answer 1

Solution:

Value of z = $\frac{x - \bar{x} }{\sigma }$

Value of z corresponding to the first group of items under 10$\%$ = 0.5 – 0.1 = 0.4

0.4 on the table = - 0.6554

Z = $\frac{x - \bar{x} }{\sigma }$

-0.6554 = $\frac{50 - \bar{x} }{\sigma }$

$-0.6554\sigma = 50 - \bar{x}$

$\bar{x}-0.6554\sigma = 50$

Value of z corresponding to the second group of items under 90$\%$ = 1 – 0.9 = 0.1

0.1 on the table = 0.5398

Z = $\frac{x - \bar{x} }{\sigma }$

0.5398= $\frac{90 - \bar{x} }{\sigma }$

$0.5398\sigma = 90 - \bar{x}$

$\bar{x} + 0.5398\sigma = 90$

Solve for the system equations:

$\bar{x}-0.6554\sigma = 50$

$\bar{x} + 0.5398\sigma = 90$

$-1.1952\sigma = -40$

$\sigma = 33.47$

The value of the standard deviation in the distribution = 33.47

Substitute the value of $\sigma$ in the first equation to derive x̅:

$\bar{x}-0.6554\sigma = 50$

$\bar{x}-0.6554(33.47) = 50$

$\bar{x} - 21.94 = 50$

$\bar{x}= 50 + 21.94 = 71.94$

$\bar{x} = 71.94$

The value of the mean of the distribution = 71.94