Question

Parts being manufactured at a plant are supposed to weigh 65 grams. Suppose the distribution of weights has a Normal distribution with mean 75 grams and a standard deviation of 22 grams. Quality control inspectors randomly select 144 parts, weigh each, and then compute the sample average weight for the 144 parts. Find The probability that the mean weight of these 144 parts is more than 80 grams or less than 70 grams is (Round to two decimals throughout and write as a % to 2 decimal places)

Answer 1

Answer:

0.64%.

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 of 75 grams and a standard deviation of 22 grams.

This means that $\mu = 75, \sigma = 22$

Sample of 144:

This means that $n = 144, s = \frac{22}{\sqrt{144}} = 1.8333$

More than 80 or less than 70:

Both are the same distance from the mean, so we find one probability and multiply by 2.

The probability that it is less than 70 is the pvalue of Z when X = 70. So

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

By the Central Limit Theorem

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

$Z = \frac{70 - 75}{1.8333}$

$Z = -2.73$

$Z = -2.73$ has a pvalue of 0.0032

2*0.0032 = 0.0064.

0.0064*100% = 0.64%

The probability is 0.64%.