M&M plain candies have weights that are normally distrbuted with a mean weight of 0.8565 gram and a standard deviation of 0.0518 gram. A random sample of 100 M&M candies is obtained from a package containing 465 candies: the package label states that the net weight is 396.9 grams. If every package has 465 candies, the mean weight of the candies must exceed 396.9/465=0.8535 for the net contents to weigh at least 3969.9 grams.
A. If 1 M&M plain cand is randomly selected, how likely is it that it weighs more than 0.8535 gram?
B. If 465 M&M plain candies are randomly selected, how likely is it that their mean weight is at least 0.8535 gram?
A. Z = (score-mean)/SD (for distribution of scores)
B. Z = (mean1 - mean2)/standard error (SE) of difference between means (for distribution of means)
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to these Z scores.
To calculate the probability in questions A and B, we need to use the concept of the standard normal distribution.
A. To find the probability that a randomly selected M&M plain candy weighs more than 0.8535 grams, we first need to standardize the value using the standard normal distribution formula.
Z = (X - μ) / σ
Where:
Z = the standardized value
X = the value we want to standardize (0.8535 grams in this case)
μ = the mean weight of M&M plain candies (0.8565 grams)
σ = the standard deviation of M&M plain candies (0.0518 grams)
Now we substitute the values and calculate Z:
Z = (0.8535 - 0.8565) / 0.0518
Calculating this, we get Z ≈ -0.0579.
In order to find the probability of a M&M plain candy weighing more than 0.8535 grams, we need to determine the area to the right of Z = -0.0579 on the standard normal distribution curve. This can be done by looking up the corresponding probability in a standard normal distribution table or using a calculator that provides the cumulative probability function.
B. To find the probability that the mean weight of 465 randomly selected M&M plain candies is at least 0.8535 grams, we need to consider the sampling distribution of the sample mean. According to the central limit theorem, as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution.
The mean weight of the sample means (μ) will be the same as the mean weight of an individual M&M plain candy (0.8565 grams), while the standard deviation (σ) of the sample means will be the standard deviation of an individual M&M plain candy divided by the square root of the sample size (sqrt(465)). Therefore:
σ = 0.0518 / sqrt(465)
To calculate the probability that the mean weight of the 465 candies is at least 0.8535 grams, we can use the standardized value formula Z:
Z = (X - μ) / (σ / sqrt(n))
Where:
X = the value we want to standardize (0.8535 grams)
μ = the mean weight of M&M plain candies (0.8565 grams)
σ = the standard deviation of an individual M&M plain candy (0.0518 grams)
n = the sample size (465)
Now we substitute the values and calculate Z:
Z = (0.8535 - 0.8565) / (0.0518 / sqrt(465))
Calculating this, we get Z ≈ -0.9722.
To find the probability of the sample mean being at least 0.8535 grams, we need to determine the area to the right of Z = -0.9722 on the standard normal distribution curve. Again, you can use a standard normal distribution table or a calculator that provides the cumulative probability function to find this probability.