Question

suppose that the mean cranial capacity for men is 1120 cc (cubic centimeters) and that the standard deviation is 300 cc. assuming that mens cranial capacities are normally distributed, complete the following statements.
(a) approximately 99.7% of men have cranial capacities between cc and cc.
(b) approximately of men have cranial capacities between 820 cc and 1420 cc.

Explanation:

Step1: Recall the empirical rule for normal distribution

The empirical rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(1\) standard deviation (\(\mu\pm\sigma\)) of the mean.
• Approximately \(95\%\) of the data lies within \(2\) standard deviations (\(\mu\pm2\sigma\)) of the mean.
• Approximately \(99.7\%\) of the data lies within \(3\) standard deviations (\(\mu\pm3\sigma\)) of the mean.

Step2: Solve part (a)

Given \(\mu = 1120\) and \(\sigma=300\).
For \(99.7\%\) of the data (within \(3\) standard deviations):
Lower bound: \(\mu - 3\sigma=1120-3\times300=1120 - 900=220\)
Upper bound: \(\mu + 3\sigma=1120+3\times300=1120 + 900=2020\)

Step3: Solve part (b)

First, find the number of standard deviations from the mean.
For \(x_1 = 820\): \(z_1=\frac{820 - 1120}{300}=\frac{- 300}{300}=-1\)
For \(x_2 = 1420\): \(z_2=\frac{1420 - 1120}{300}=\frac{300}{300}=1\)
Since the data lies within \(1\) standard deviation (\(\mu\pm\sigma\)) of the mean, by the empirical rule, approximately \(68\%\) of the data lies in this range.

Answer:

(a) \(220\) cc and \(2020\) cc.
(b) \(68\%\)