Given : The iq of students in a CERTAIN COLLEGE is assumed to be normally distributed with mean 100 and variance 25 . 3 students are selected at random

To FIND :  probability that all of them have iq between 102 and 110.

Step-by-step explanation:

mean 100

Variance = 25

Variance = (Standard deviation)²

=> (Standard deviation)² = 25

=> Standard deviation = 5

Mean = 100

SD = 5

Z score = ( Value - Mean)/SD

Z = (102 - 100)/5

=>  Z = 0.4

A(0.4)=0.4772  => 47.72 % data lies above 102

Z = (110 - 100)/5

=>  Z = 2

A(2)=0.1554 => 15.54 % data lies above 110

Data lies in between 102 & 110

= 47.72 - 15.54

= 32.18 %

0.3218  is the probability that a students selected have score between 102 & 110

now probability all three students selected have score between 102 & 110

= (0.3218)³

= 0.033

0.033 is the probability that all 3 students have score between 102 & 110

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