A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 245.7-cm and a standard deviation of 1.8-cm. For shipment, 5 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 245.1-cm and 248.2-cm.
P(245.1-cm < M < 248.2-cm) =
Solution :
Given that,
mean = = 245.7
standard deviation = = 1.8
n = 5
= = 245.7
= / n = 1.8 / 5 = 0.805
P(245.1 < M < 248.2)
= P[(245.1 - 245.7) / 0.805 < (M - ) / < (248.2 - 245.7) / 0.805)]
= P(-0.745 < Z < 3.106)
= P(Z < 3.106 ) - P(Z < -0.745)
Using z table,
= 0.9991 - 0.2281
= 0.7710
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