IQs are known to be normally distributed with mean 100 and standard deviation 15. In a...

IQs are known to be normally distributed with mean 100 and standard deviation 15. In a random sample of 33 people, find the probability that the average IQ is between 96 and 104.

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Expert Solution

Given that ,

mean = = 100

standard deviation = = 15

n = 33

= 100

= / $\sqrt$ n= 15 / $\sqrt$ 33=2.61

P(96< < 104) = P[(96-100) /2.61 < ( - ) / < (104-100) /2.61 )]

= P(-1.53 < Z <1.53 )

= P(Z <1.53) - P(Z < -1.53)

Using z table

=0.937 - 0.0630

probability= 0.8740

orchestra answered 1 year ago

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