Question

A certain brand of candies have a mean weight of 0.8616g and a standard deviation of 0.0518 based on the sample of a package containing 447 candies. The package label stated that the net weight is 381.8g If every package has 447 candies the mean weight of the candies must exceed 381.8/447=0.8542 for the net contents to weight at least 381.8g.
a) If 1 candy is reandomly selected, find the probability that it weights more than 0.8542g.
The probability is ________.
(round to four decimal places as needed)
b) If 447 candies are reandomly selected find the probability that their mean weight is at least 0.8542 g.
the probability that a sample of 447 candies will have a mean of 0.8542g or greater is __________.
(round to four decimal places as needed)
c) Given these results does it seem that the candy company is providing consumers with the amount claimed on the label?

Answer 1

Answer:

a) The probability is 0.5557 = 55.57%.

b) The probability that a sample of 447 candies will have a mean of 0.8542g or greater is 0.9987 = 99.87%.

c) Yes, because there is a very large probability, of 99.87%, that the amount will be at least the one claimed on the label.

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 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 $s = \frac{\sigma}{\sqrt{n}}$.

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

Mean weight of 0.8616g and a standard deviation of 0.0518

This means that $\mu = 0.8616, \sigma = 0.0518$

a) If 1 candy is reandomly selected, find the probability that it weights more than 0.8542g.

This is 1 subtracted by the pvalue of Z when X = 0.8542. So

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

$Z = \frac{0.8542 - 0.8616}{0.0518}$

$Z = -0.14$

$Z = -0.14$ has a pvalue of 0.4443

1 - 0.4443 = 0.5557

The probability is 0.5557 = 55.57%.

b) If 447 candies are reandomly selected find the probability that their mean weight is at least 0.8542 g.

Sample of 447 means that $n = 447, s = \frac{0.0518}{\sqrt{447}} = 0.00245$

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

By the Central Limit Theorem

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

$Z = \frac{0.8542 - 0.8616}{0.0245}$

$Z = -3.02$

$Z = -3.02$ has a pvalue of 0.0013

1 - 0.0013 = 0.9987

The probability that a sample of 447 candies will have a mean of 0.8542g or greater is 0.9987 = 99.87%.

c) Given these results does it seem that the candy company is providing consumers with the amount claimed on the label?

Yes, because there is a very large probability, of 99.87%, that the amount will be at least the one claimed on the label.