Question

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 55 ounces and a standard deviation of 6 ounces.

Use the Standard Deviation Rule, also known as the Empirical Rule.
Suggestion: sketch the distribution in order to answer these questions.
a) 95% of the widget weights lie between _______ and ________.
b) What percentage of the widget weights lie between 37 and 67 ounces?c) What percentage of the widget weights lie above 49 ?

Answer 1

a) For this case and using the empirical rule we can find the limits in order to have 9% of the values:

`\mu -2\sigma = 55 -2*6 =43`

`\mu +2\sigma = 55 +2*6 =67`

95% of the widget weights lie between 43 and 67

b) For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

`100 -0.15-2.5 = 97.85`

c) We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

Explanation:

For this case our random variable of interest for the weights is bell shaped and we know the following parameters.

`\mu = 55, \sigma =6`

We can see the illustration of the curve in the figure attached. We need to remember that from the empirical rule we have 68% of the values within one deviation from the mean, 95% of the data within 2 deviations and 99.7% of the values within 3 deviations from the mean.

Part a

For this case and using the empirical rule we can find the limits in order to have 9% of the values:

`\mu -2\sigma = 55 -2*6 =43`

`\mu +2\sigma = 55 +2*6 =67`

95% of the widget weights lie between 43 and 67

Part b

For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

`100 -0.15-2.5 = 97.85`

Part c

We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%