A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 129.2-cm and a standard deviation of 0.5-cm. For shipment, 27 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 129.3-cm. P(M > 129.3-cm) = __________
Solution :
Given that,
mean = = 129.2
standard deviation = = 0.5
n=27
= =129.2
= / n = 0.5/ 27 = 0.096
P(M > 129.3) = 1 - P(M <129.3 )
= 1 - P[( - ) / < (129.3-129.2) /0.096 ]
= 1 - P(z <1.04 )
Using z table
= 1 - 0.8508
probability= 0.1492
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