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A worldwide organization of academics claims that the mean IQ score of its members is 112, with a standard deviation of 15. A

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Solution :

Given that ,

mean = \mu = 112

standard deviation = \sigma = 15

n = 35

\mu\bar x = 112

\sigma\bar x =\sigma  / \sqrt n = 15/ \sqrt 35=2.535

P(\bar x <114.4 ) = P[(\bar x - \mu \bar x ) / \sigma \bar x < (114.4-112) / 2.535]

= P(z < 0.95)

Using z table

probability=  0.8289

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