Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 177.5-cm and a standard deviation of 1.2-cm. For shipment, 7 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.

Answer 1

Answer: the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm is 0.96

Explanation:

Since the lengths of the steel rods are normally distributed, then according to the central limit theorem,

z = (x - µ)/(σ/√n)

Where

x = sample mean lengths of the steel rods

µ = population mean length of the steel rods

σ = standard deviation

n = number of samples

From the information given,

µ = 177.5 cm

x = 178.3 cm

σ = 1.2 cm

n = 7

the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm is expressed as

P(x < 178.3)

Therefore,

z = (178.3 - 177.5)/(1.2/√7) = 1.76

Looking at the normal distribution table, the probability corresponding to the z score is 0.96

Therefore,

P(x < 178.3) = 0.96

Answer 2

96.08% probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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.

In this problem, we have that:

`\mu = 177.5, \sigma = 1.2, n = 7, s = (1.2)/(√(7)) = 0.4536`

Find the probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.

This is the pvalue of Z when X = 178.3. So

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

By the Central Limit Theorem

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

`Z = (178.3 - 177.5)/(0.4536)`

`Z = 1.76`

`Z = 1.76` has a pvalue of 0.9608

96.08% probability that the average length of a randomly selected bundle of steel rods is less than 178.3-cm.