A machine at Katz Steel Corporation makes 5-inch-long nails. The probability distribution of the lengths of these nails is approximately normal with a mean of 5 inches and a standard deviation of 0.10 inch. The quality control inspector takes a sample of 16 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than 4.94 inches or greater than 5.06 inches, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 16 nails, the inspector will conclude that the machine needs an adjustment?
Round your answer to 4 decimal places.
Probability =
Solution :
Given that,
mean = = 5
standard deviation = = 0.10
_{} = / n = 0.10 / 16 = 0.025
=1 - P[(4.94 - 5) / 0.025< ( - _{}) / _{} < (5.06 - 5) / 0.025)]
= 1 - P(-2.4 < Z < 2.4)
=1 - P(Z < 2.4) - P(Z < -2.4)
= 1 - P(0.9918 - 0.0082)
= 1 - 0.9836
= 0.0164
probability = 0.0164
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