A company produces steel rods. The lengths of the steel rods are
normally distributed with a mean of 211.4-cm and a standard
deviation of 1.3-cm. For shipment, 5 steel rods are bundled
Find the probability that the average length of a randomly selected bundle of steel rods is greater than 211.5-cm.
P(M > 211.5-cm) =
mean = = 211.4
standard deviation = = 1.3
= / n = 1.3 / 5 = 0.58
P(M > 211.5) = 1 - P(M <211.5 )
= 1 - P[(M - ) / < (211.5 - 211.4) / 0.58]
= 1 - P(z <0.17)
Using z table
= 1 - 0.5675
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