Question

Weights measured in grams of randomly selected M&M plain candies:

Answer 1

Answer:

The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Sample mean:

$\overline{x} = \frac{0.957 + 0.912 + 0.925 + 0.886 + 0.920 + 0.958 + 0.915 + 0.914 + 0.947 + 0.939 + 0.842}{11} = 0.9195$

Sample standard deviation:

$s = \sqrt{\frac{(0.957-0.9195)^2 + (0.912-0.9195)^2 + (0.925-0.9195)^2 + (0.886-0.9195)^2 + ...}{10}} = 0.0336$

What is the range of weights of the middle 99.7% of M&M’s?

By the Empirical Rule, within 3 standard deviations of the mean, so:

0.9195 - 3*0.0336 = 0.8187.

0.9195 + 3*0.0336 = 1.0203.

The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.