Question

1. your favorite bag of candy says that it should have 8oz of the candy in the bag. you weigh the contents and find that you have 7.8oz. you log onto the candys website to write a strongly worded email, and see that they have a disclaimer that the weights of the bags will vary according to a normal distribution with a mean weight of 8oz and a standard deviation of 0.25oz. which of the following is an appropriate conclusion based on this information? (note: anything with a z - score greater than 3 or less than negative 3 would be considered outliers, outliers would be considered unreasonable) (a) 7.8oz is still a reasonable weight of your candy bag, since it falls 0.8oz below the mean. (b) 7.8oz is still a reasonable weight of your candy bag, since it falls 0.8 standard deviations below the mean. (c) 7.8 is not a reasonable weight of your candy bag, since it falls below the mean. (d) 7.8 is not a reasonable weight of your candy bag, since it falls above the mean. (e) none of the above are appropriate conclusions.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 7.8$, $\mu=8$, and $\sigma = 0.25$.
$z=\frac{7.8 - 8}{0.25}=\frac{- 0.2}{0.25}=-0.8$

Step2: Analyze the z - score

Since $- 3< - 0.8<3$, the value is not an outlier. A z - score of - 0.8 means the value of 7.8 oz is 0.8 standard deviations below the mean.

Answer:

B. 7.8oz is still a reasonable weight of your candy bag, since it falls 0.8 standard deviations below the mean.