Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 219.7-cm and a standard deviation of 1.6-cm. For shipment, 21 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 219.7-cm and 220-cm.

Answer 1

Answer: The probability that the average length of a randomly selected bundle of steel rods is between 219.7-cm and 220-cm is 0.31

Explanation:

Since the lengths of the steel rods are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = lengths of walleye fishes.

µ = mean length

σ = standard deviation

From the information given,

µ = 219.7-cm

σ = 1.6-cm

The probability that the average length of a randomly selected bundle of steel rods is between 219.7-cm and 220-cm is expressed as

P(219.7 ≤ x ≤ 220)

For x = 219.7,

z = (219.7 - 219.7)/1.6/√21 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

For x = 220,

z = (220 - 219.7)/1.6/√21 = 0.86

Looking at the normal distribution table, the probability corresponding to the z score is 0.81

P(219.7 ≤ x ≤ 220) = 0.81 - 0.5 = 0.31

Answer 2

30.51% probability that the average length of a randomly selected bundle of steel rods is between 219.7-cm and 220-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 sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 219.7 \sigma = 1.6, n = 21, s = (1.6)/(√(21)) = 0.3491`

Find the probability that the average length of a randomly selected bundle of steel rods is between 219.7-cm and 220-cm.

This is the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 219.7.

X = 220

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

By the Central Limit Theorem

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

`Z = (220 - 219.7)/(0.3491)`

`Z = 0.86`

`Z = 0.86` has a pvalue of 0.8051

X = 219.7

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

`Z = (219.7 - 219.7)/(0.3491)`

`Z = 0`

`Z = 0` has a pvalue of 0.5

0.8051 - 0.5 = 0.3051

30.51% probability that the average length of a randomly selected bundle of steel rods is between 219.7-cm and 220-cm.