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 6.13 ounces and a standard deviation of 0.15 ounce. Suppose that you draw a random sample of 32 cans. Find the probability that the mean weight of the sample is less than 6.11 ounces. Probability =

Answer 1

22.66% probability that the mean weight of the sample is less than 6.11 ounces.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, 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 = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

`\mu = 6.13, \sigma = 0.15, n = 32, s = (0.15)/(√(32)) = 0.0265`

Find the probability that the mean weight of the sample is less than 6.11 ounces.

This is the pvalue of Z when X = 6.11.

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

By the Central Limit Theorem

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

`Z = (6.11 - 6.13)/(0.0265)`

`Z = -0.75`

`Z = -0.75` has a pvalue of 0.2266

22.66% probability that the mean weight of the sample is less than 6.11 ounces.