Mathematics, 28.07.2021 17:30, awgggg

The manufacturer of cans of salmon that are supposed to have a net weight of 6 ounces tells you that the net weight is actually a normal random variable with a mean of 6.05 ounces and a standard deviation of .18 ounces. Suppose that you draw a random sample of 36 cans. a. Find the probability that the mean weight of the sample is less than 5.97 ounces.
b. Suppose your random sample of 36 cans of salmon produced a mean weight that is less than 5.97 ounces. Comment on the statement made by the manufacturer.

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answered: jaylan11brown

a) 0.0038 = 0.38% probability that the mean weight of the sample is less than 5.97 ounces.

b) Given a mean of 6.05 ounces, it is very unlikely that a sample mean of less than 5.97 ounces, which means that the true mean must be recalculated.

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 establishes 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 6.05 ounces and a standard deviation of .18 ounces.

This means that $\mu = 6.05, \sigma = 0.18$

Sample of 36:

This means that $n = 36, s = \frac{0.18}{\sqrt{36}} = 0.03$

a. Find the probability that the mean weight of the sample is less than 5.97 ounces.

This is the p-value of z when X = 5.97. So

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

By the Central Limit Theorem

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

$Z = \frac{5.97 - 6.05}{0.03}$

Z = -2.67

Z = -2.67 has a p-value of 0.0038.

0.0038 = 0.38% probability that the mean weight of the sample is less than 5.97 ounces.

b. Suppose your random sample of 36 cans of salmon produced a mean weight that is less than 5.97 ounces. Comment on the statement made by the manufacturer.

Given a mean of 6.05 ounces, it is very unlikely that a sample mean of less than 5.97 ounces, which means that the true mean must be recalculated.

answered: Guest

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