The quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a variance of 100. If he is correct, what is the probability that the mean of a sample of 57 computers would differ from the population mean by less than 0.9 months? Round your answer to four decimal places.
We have to calculate
P( | - | < 0.9 ) = ?
P( | - 120 | < 0.9 ) = ?
P(-0.9 < - 120 < 0.9) = P(119.1 < < 120.9)
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 119.1 - 120 ) / ( 10 / √(57))
Z = -0.68
Z = ( 120.9 - 120 ) / ( 10 / √(57))
Z = 0.68
P ( 119.1 < X̅ < 120.9 ) = P ( Z < 0.68 ) - P ( Z <
-0.68 )
P ( 119.1 < X̅ < 120.9 ) = 0.7517 - 0.2483
P ( 119.1 < X̅ < 120.9 ) = 0.5035
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