Question

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.

Answer 1

The probability that the mean weight of a sample of 36 cans is less than 5.97 ounces is 0.38%, calculated using the Central Limit Theorem. A sample mean of less than 5.97 ounces suggests potential inconsistency with the manufacturer's claim of 6.05 ounces mean weight.

Step-by-step explanation:

To answer the student's question regarding the probability of the mean weight of a sample of 36 cans of salmon being less than 5.97 ounces, we can use the Central Limit Theorem. Given that the mean weight of the cans is normally distributed with a mean (μ) of 6.05 ounces and a standard deviation (σ) of 0.18 ounces, the sampling distribution of the sample mean for a sample size (n) of 36 will also be normally distributed with a mean equal to the population mean (μ = 6.05 ounces) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (standard error, σ/√n = 0.18/√36 = 0.03 ounces).

To find the probability that the sample mean is less than 5.97 ounces, we calculate the z-score of 5.97 using the formula:

Z = (X - μ) / (σ/√n)

Inserting the given values:

Z = (5.97 - 6.05) / 0.03 = -2.67

Using standard normal distribution tables or a calculator, we find that the probability of a z-score being less than -2.67 is approximately 0.0038. Hence, there's a 0.38% chance that the sample mean is less than 5.97 ounces.

Regarding part b, if the sample mean is less than 5.97 ounces, it suggests that there may be an inconsistency or issue with the manufacturer's statement. The expected mean weight is claimed to be 6.05 ounces, so a sample mean significantly lower than this may imply that the cans are underfilled or that there is variability in the filling process that the manufacturer should investigate.

Answer 2

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 = (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 = (\sigma)/(√(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 = (0.18)/(√(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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.