Question

Suppose a batch of steel rods produced at a steel plant have a mean length of 150 millimeters, and a variance of 100. If 275 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by greater than 0.57 millimeters? Round your answer to four decimal places. Answer

Answer 1

Answer:

0.3446

Step-by-step 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(which is the square root of the variance) $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 150, \sigma = \sqrt{100} = 10, n = 275, s = \frac{10}{\sqrt{275}} = 0.6030$

If 275 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by greater than 0.57 millimeters?

Either it differs by 0.57 millimeters or less, or it differs by more than 0.57 millimeters. The sum of the probabilities of these events is decimal 1.

Probability it differs by 0.57 millimeters or less.

pvalue of Z when X = 150+0.57 = 150.57 subtracted by the pvalue of Z when X = 150-0.57 = 149.43.

X = 150.57

$Z = \frac{X - \mu}{\sigma}$

By the Central limit theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{150.57 - 150}{0.6030}$

$Z = 0.945$

$Z = 0.945$ has a pvalue of 0.8277

X = 149.43

$Z = \frac{X - \mu}{s}$

$Z = \frac{149.43 - 150}{0.6030}$

$Z = -0.945$

$Z = -0.945$ has a pvalue of 0.1723

0.8277 - 0.1723 = 0.6554

0.6554 probabiliity it differs by 0.57 millimeters or less

Probability it differs by more than 0.57 millimeters

1 - 0.6554 = 0.3446

0.3446 = 34.46% probability that the mean length of the sample rods would differ from the population mean by greater than 0.57 millimeters.