Question

1/ The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 63 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) 68% of the widget weights lie between  and

b) What percentage of the widget weights lie between 45 and 69 ounces?  %

c) What percentage of the widget weights lie above 51 ?  %

2/ dult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the height of a man with a z-score of 2.9 (to 2 decimal places)

Answer 1

Here,

mean = 63

standard deviation = 6

Using empirical 68 - 95 - 99.7 rule, we can write

68% values of a bell shaped distribution will lie between the interval (mean - standard deviation, mean + standard deviation)

95% values of a bell shaped distribution will lie between the interval (mean - 2*standard deviation, mean + 2*standard deviation)

99.7% values of a bell shaped distribution will lie between the interval (mean - 3*standard deviation, mean + 3*standard deviation)

Now,

a) 68% of the widget weights lie between (mean - standard deviation, mean + standard deviation) = (63 - 6, 63 + 6) = (57, 69)

b) 99.7% values of a bell shaped distribution will lie between the interval (mean - 3*standard deviation, mean + 3*standard deviation) = (63 - 3*6, 63 + 3.*6) = (45, 81)

Percent of values less than 45 = (100-99.7)/2 % = 0.15%

68% of the widget weights lie between (57, 69)

Percent of values greater than 69 = (100 - 68)/2 % = 16%

Percentage of values less than 45 and greater than 69 = (0.15 + 16)%= 16.15%

So, the percentage of the widget weights lie between 45 and 69 ounces = (100 - 16.15)% = 83.85%

c)  68% of the widget weights lie between (mean - 2*standard deviation, mean + 2*standard deviation) = (63 -2*6, 63 + 2*.6)

= (51, 75)

The percent of values lie outside the interval (51, 75) = (100 - 95)% = 5%

The percent of values less than 51 = 5/2 % = 2.5%

The percentage of the widget weights lie above 51 = (100 - 2.5)% = 97.5%

2. The height of a man with a z-score of 2.9 is = mean + (z-score*standard deviation) = 69 + (2.9*2.8) = 77.12