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There is a test which is normally distributed with a mean of 30 and a standard...

There is a test which is normally distributed with a mean of 30 and a standard deviation of 4. Find the probability that a sample of 36 scores will have a mean at most 31.5.

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Expert Solution

Solution :

Given that ,

mean = = 30

standard deviation = = 4

n = 36

=    = 30

= / n = 4 / 36 = 0.667

P( 31.5 ) = P(( - ) / (31.5 - 30) / 0.667 )

= P(z 2.25)

Using z table

= 0.9878

orchestra answered 2 years ago
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