Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 109-cm and a standard deviation of 1.6-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 less than 108.7-cm. P(M < 108.7-cm) =

Answer 1

The probability that the average length of a randomly selected bundle of steel rods is 0.165

Given population mean(μ)=109cm

standard deviation(σ)=1.6cm

sample size(n)=27

Let x represents the lengths of an individual rod then the z-score for

x can be given as:

z=x-μ/σ----------(1)

Also, the z score for the sample mean, x:

z=(¯x−μ)/σx

The standard error:

σ¯x=σ/√n-------------(2)

      =1.6/√27

       =1.6/5.196

σ¯x =0.3079

For a bundle of 27rods, the probability that the average length is between 109 cm and 108.7-cm:

considering m=108.7cm

P(n-m/σx)------------(3)

$P(109 < x < 108.7)=P(\frac{109-108.7}{0.3079} )\\\\ P(109 < x < 108.7)=P(z < -0.974)$

                              =0.165.

And also find using another method:

P(z < (M - population mean)/(standard deviation/square root(sample size))

P(z < (108.7-109)/(1.6/sqrt(27)) = P(z < -0.974) = 0.165.

To know more about the probability of average length:

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