Question

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) =

Answer #1

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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