Question

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) = __________

Answer #1

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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