Question #212300

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.


1
Expert's answer
2021-07-01T13:18:16-0400

Solution:

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

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

0.6554σ=50xˉ-0.6554\sigma = 50 - \bar{x}

xˉ0.6554σ=50\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 = xxˉσ\frac{x - \bar{x} }{\sigma }

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

0.5398σ=90xˉ0.5398\sigma = 90 - \bar{x}

xˉ+0.5398σ=90\bar{x} + 0.5398\sigma = 90

Solve for the system equations:

xˉ0.6554σ=50\bar{x}-0.6554\sigma = 50

xˉ+0.5398σ=90\bar{x} + 0.5398\sigma = 90

1.1952σ=40-1.1952\sigma = -40

σ=33.47\sigma = 33.47

The value of the standard deviation in the distribution = 33.47


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

xˉ0.6554σ=50\bar{x}-0.6554\sigma = 50

xˉ0.6554(33.47)=50\bar{x}-0.6554(33.47) = 50

xˉ21.94=50\bar{x} - 21.94 = 50

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

xˉ=71.94\bar{x} = 71.94

The value of the mean of the distribution = 71.94

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