Question

The amount spent for goods online is normally distributed with mean of $125 and a Standard
deviation of $25 for a certain age group.
i. what the percent spent more than $175
ii. what percent spent between $100 and $150
iii. what is the probability that they spend more than $50

Answer 1

Given : The amount spent for goods online is normally distributed with mean of $125 and a Standard deviation of $25 for a certain age group.

To Find : i. what the percent spent more than $175

ii. what percent spent between $100 and $150

iii. what is the probability that they spend more than $50

Solution:

Mean = 125

Standard Deviation SD  = 25

Z score =  (Value - Mean)/SD

spent more than $175

=> Z score = ( 175 - 125)/25

=> Z score = 2

Z score 2 = 97.72 %

spent more than $175  = 100 -  97.72 = 2.28 %

2.28 % more than $175

Spent between  100 & 150

Z score for 150 = (150 - 125)/25  = 1    

Z Score 1  = 84.13 %

Z score for 100 = (100 - 125)/25  = -1    

Z Score -1  = 15.87 %

=>  percent spent between $100 and $150  = 84.13 - 15.87 =  68.26 %

68.26 % spent between $100 and $150

Z score for 50 = (50 - 125)/25  = - 3

= 0.13 % => Spend more than  = 100 - 0.13% = 99.87 %

0.9987 is the probability that  they spend more than $50

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