Question

Each day a manufacturing plant receives a large shipment of drums of Chemical ZX-900. These drums are supposed to have a mean fill of 50 gallons, while the fills have a standard deviation known to be 0.6 gallon. Suppose the mean fill for the shipment is actually 50 gallons. If we draw a random sample of 100 drums from the shipment, what is the probability that the average fill for the 100 drums is between 49.88 and 50.12 gallons

Answer 1

Answer:

0.9544 = 95.44% probability that the average fill for the 100 drums is between 49.88 and 50.12 gallons.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, 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.

Suppose the mean fill for the shipment is actually 50 gallons. Standard deviation of 0.6 gallons.

This means that $\mu = 50, \sigma = 0.6$

Sample of 100:

This means that $n = 100, s = \frac{0.6}{\sqrt{100}} = 0.06$

What is the probability that the average fill for the 100 drums is between 49.88 and 50.12 gallons?

This is the pvalue of Z when X = 50.12 subtracted by the pvalue of Z when X = 49.88. So

X = 50.12

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

By the Central Limit Theorem

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

$Z = \frac{50.12 - 50}{0.06}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 49.88

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

$Z = \frac{49.88 - 50}{0.06}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the average fill for the 100 drums is between 49.88 and 50.12 gallons.