Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 193.5-cm and a standard deviation of 2.1-cm. For shipment, 10 steel rods are bundled together. Round all answers to four decimal places if necessary. What is the distribution of X X? X X ~ N(,) What is the distribution of ¯ x x¯? ¯ x x¯ ~ N(,) For a single randomly selected steel rod, find the probability that the length is between 193.8-cm and 194.2-cm. For a bundled of 10 rods, find the probability that the average length is between 193.8-cm and 194.2-cm. For part d), is the assumption of normal necessary? NoYes

Answer 1

a) X~N(193.5,2.1)

b) x¯ ~N(193.5,0.6641)

c) 0.0736 = 7.36% probability that the length is between 193.8-cm and 194.2-cm.

d) 0.1795 = 17.95% probability that the average length is between 193.8-cm and 194.2-cm.

e) No, since the population is normally distributed.

Explanation:

To solve this question, we need to understand the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a) What is the distribution of X?

`\mu = 193.5, \sigma = 2.1`.

So

X~N(193.5,2.1)

b) What is the distribution of x¯?

Sample,
`n = 10, s = (2.1)/(√(10)) = 0.6641`

x¯ ~N(193.5,0.6641)

c) For a single randomly selected steel rod, find the probability that the length is between 193.8-cm and 194.2-cm.

This is the pvalue of Z when X = 194.2 subtracted by the pvalue of Z when X = 193.8.

X = 194.2

`Z = (X - \mu)/(\sigma)`

`Z = (194.2 - 193.5)/(2.1)`

`Z = 0.33`

`Z = 0.33` has a pvalue of 0.6293

X = 193.8

`Z = (X - \mu)/(\sigma)`

`Z = (193.8 - 193.5)/(2.1)`

`Z = 0.14`

`Z = 0.14` has a pvalue of 0.5557

0.6293 - 0.5557 = 0.0736

0.0736 = 7.36% probability that the length is between 193.8-cm and 194.2-cm.

d) For a bundled of 10 rods, find the probability that the average length is between 193.8-cm and 194.2-cm.

Same as above, just that now we use the standard deviation for the sample means.

X = 194.2

`Z = (X - \mu)/(s)`

`Z = (194.2 - 193.5)/(0.6641)`

`Z = 1.05`

`Z = 1.05` has a pvalue of 0.8531

X = 193.8

`Z = (X - \mu)/(s)`

`Z = (193.8 - 193.5)/(0.6641)`

`Z = 0.45`

`Z = 0.45` has a pvalue of 0.6736

0.8531 - 0.6736 = 0.1795

0.1795 = 17.95% probability that the average length is between 193.8-cm and 194.2-cm.

e) For part d), is the assumption of normal necessary?

By the Central Limit Theorem, this assumption is only necessary if the underlying population is not normally distributed. Since the lengths of the steel rods are normally distributed, this assumption is not necessary.