The breaking strength of a brand of safety belt is known to

The breaking strength of a brand of safety belt is known to be approximately normally distributed with a population mean of 8,000 pounds and a population standard deviation of 250 pounds. If a simple random sample of 25 belts in chosen, what is the probability that the sample mean will differ from the true mean by more than 100 pounds in either direction? Please show calculations.

Solution

Probability of sample mean differ by more than 100 pounds from true mean

= P(X<7900) and P(X>8100)

= 1 - P(7900 < X < 8100)

= 1 - P(X<8100) - P(X<7900)

= 1 - P(Z < 8100 - 8000/250/sqrt(25)) - P(Z<7900 - 8000/250/sqrt(25))

= 1 - P(Z <2) + P(Z < -2)

= 1 -0.9772 + 0.0228

= 0.0456