Question

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 0.12 inch. The quality control inspector takes a sample of 36 nails once a week and calculates the mean length of these nails. If the mean of this sample is either less than 4.95 inches or greater than 5.05 inches, the inspector concludes that the machine needs an adjustment. What is the probability that based on a sample of 36 nails, the inspector will conclude that the machine needs an adjustment?

Round your answer to 4 decimal places.

Probability =

Answer #1

Here, μ = 5, σ = 0.02, x1 = 4.95 and x2 = 5.05. We need to compute P(4.95<= X <= 5.05). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ

z1 = (4.95 - 5)/0.02 = -2.5

z2 = (5.05 - 5)/0.02 = 2.5

Therefore, we get

P(4.95 <= X <= 5.05) = P((5.05 - 5)/0.02) <= z <= (5.05
- 5)/0.02)

= P(-2.5 <= z <= 2.5) = P(z <= 2.5) - P(z <=
-2.5)

= 0.9938 - 0.0062

= 0.9876

probability = 1 - 0.9876 = 0.0124

Ans : 0.0124

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