The probability that the mean weight of the sample is less than 6.02 ounces is approximately 0.0021 or 0.21%.
To find the probability that the mean weight of the sample is less than 6.02 ounces, you can use the central limit theorem and approximate the distribution of the sample mean using the normal distribution since the sample size is sufficiently large (n > 30).
Given:
Population mean (μ) = 6.07 ounces
Population standard deviation (σ) = 0.1 ounce
Sample size (n) = 33
First, find the standard error of the mean (SE), which is the standard deviation of the sample mean:
Where:
σ = Population standard deviation
n = Sample size
Now, find the z-score for the sample mean of 6.02 ounces using the formula:
z=
z=
z≈
≈−2.88
To find the probability that the sample mean is less than 6.02 ounces, calculate the probability using the z-table or a calculator with the standard normal distribution.
The probability can be found by looking up the z-score of -2.88 in a standard normal distribution table or using a calculator, which will give you the area to the left of this z-score.
Consulting a standard normal distribution table or using a calculator, the probability that the mean weight of the sample is less than 6.02 ounces is approximately 0.0021 or 0.21%.
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 6.07 ounces and a standard deviation of 0.1 ounce. Suppose that you draw a random sample of 33 cans. Find the probability that the mean weight of the sample is less than 6.02 ounces.