The iq of students in a certain college is assumed to be normally distributed with mean 100 and variance 25. If 3 students are selected at random, find the probability that all of them have iq between 102 and 110. (Given A(0.4)=0.4772 and A(2)=0.1554)
Answers
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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