) The IQ of students in a certain college is assumed to be normally distributed with mean 100 and variance 25. If two students are selected at random , find the probability that (i) both of them have IQ between 102 and 110 (ii) at least one of them have IQ between 102 and 110 (iii) at most one of them have IQ between 102 and 110.
Answers
Given : The IQ of students in a certain college is assumed to be normally distributed with mean 100 and variance 25. Two students are selected at random
To find : probability that (i) both of them have IQ between 102 and 110 (ii) at least one of them have IQ between 102 and 110 (iii) at most one of them have IQ between 102 and 110.
Solution:
Mean = 100
Variance = 25
SD (Standard Deviation = √Variance = √25 = 5
Z Score = (Value - Mean)/SD
IQ Between 102 & 110
Z score for 102 = (102 - 100)/5 = 0.4 => 0.6554
Z score for 110 = (110 - 100)/5 = 2 => 0.9772
=> Probability of Score between 102 & 110 = 0.9772 - 0.6554
= 0.3218
probability that both of them have IQ between 102 and 110 = (0.3218)² = 0.1036
probability that at least one of them have IQ between 102 and 110 = 1 - none of them
= 1 - ( 1 - 0.3218)²
= 1 - (0.6782)²
= 0.54
probability that at most one of them have IQ between 102 and 110.
= 1 - both of them
= 1 - 0.1036
= 0.8964
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