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Scores on a standard test of mechanical aptitude are normally distributed with a mean of 72 and a s. d. of 12. If 36 subjects

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Solution :

Given that ,

mean = \mu = 72

standard deviation = \sigma = 12

n = 36

\mu\bar x = 72

\sigma\bar x = \sigma / \sqrt n = 12/ \sqrt36 = 2

P(\bar x > 69) = 1 - P(\bar x <69 )

= 1 - P[( \bar x - \mu \bar x ) / \sigma \bar x < (69-72) /2 ]

= 1 - P(z <-1.5 )

Using z table

= 1 - 0.0668

= 0.9332

probability= 0.933

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