Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.
Find the percentage of the population which has an IQ score between 104 and 112.
Solution :
Given that ,
mean =
= 100
standard deviation =
= 15
P(104< x <112 ) = P[(104-100) / 15< (x -
) /
< (112-100) / 15)]
= P( 0.27< Z <0.8 )
= P(Z < 0.8) - P(Z <0.27 )
Using z table
= 0.7881-0.6064
= 0.1817
answer=18.17%
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