Question

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
together.

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) =

Answer #1

Solution :

Given that,

mean = = 211.4

standard deviation = = 1.3

n=5

_{}
=
=211.4

_{}
=
/
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

probability= 0.4325

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