Question

Lloyd's Cereal company packages cereal in 1 pound boxes (16 ounces). A sample of 49 boxes is selected at random from the production line every hour, and if the average weight is less than 15 ounces, the machine is adjusted to increase the amount of cereal dispensed. If the mean for 1 hour is 1 pound and the standard deviation is 0.2 pound, what is the probability that the amount dispensed per box will have to be increased

Answer 1

Answer:

0.0143 = 1.43% probability that the amount dispensed per box will have to be increased

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.

Mean for 1 hour is 1 pound and the standard deviation is 0.2 pound

This means that $\mu = 1, \sigma = 0.2$

Sample of 49:

This means that $n = 49, s = \frac{0.2}{\sqrt{49}} = 0.0286$

What is the probability that the amount dispensed per box will have to be increased?

This is the probability of the sample mean being less than 15 pounds = 15/16 = 0.9375 ounces, which is the pvalue of Z when X = 0.9375.

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

By the Central Limit Theorem

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

$Z = \frac{0.9375 - 1}{0.0286}$

$Z = -2.19$

$Z = -2.19$ has a pvalue of 0.0143

0.0143 = 1.43% probability that the amount dispensed per box will have to be increased