Step 1

Given:

For a sample size of 277, the sample mean waist circumference of 18-year-old American males, was 86.3 cm. The percentile values are also provided.

 
Step 2

a.

Required to find is it plausible that the waist size distribution is at least approximately normal.

The estimated mean waist circumference of 18-year-old American males, is the 50th percentile which is given as 81.3 cm. Also, the sample mean waist circumference of 18-year-old American males, is 86.3 cm.

The curve of the normal distribution is symmetric about its mean, and mean, median and mode of normal distribution coincides.

Here it can be seen that the mean (86.3) is greater than the median (81.3).

Thus, it is not plausible that the waist size distribution is at least approximately normal.

The shape of the population distribution can be conjecture using the relation between mean, median and mode depending upon skewness.

1. When mean = median = mode, the distribution is symmetric.

2. When mean > median > mode, the distribution is positively skewed.

3. When mean < median < mode, the distribution is negatively skewed.

Here, mean (86.3) > median (81.3).

Therefore, waist size distribution is said to be positively skewed.

 
Step 3

b.

Suppose that the population mean waist size is 85 cm and that the population standard deviation is 15 cm. Required to find how likely is it that a random sample of 277 individuals will result in a sample mean waist size of at least 86.3 cm.

The population mean ($\mu$) waist size is given as 85 cm and that the population standard deviation ($\sigma$) is 15 cm. Also, the sample size (n) is 277.

Let $\bar X$ be the sample mean and ${\sigma _{\bar X}}$ be the sample standard deviation of the waist sizes.

The mean of $\bar X$ is 85 as it is unaffected by sample size.

The sample standard deviation is obtained as:

\[\begin{array}{c} {\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }}\\ = \frac{{15}}{{\sqrt {277} }}\\ = \frac{{15}}{{16.6}}\\ = 0.9 \end{array}\]

As the sample size (277) is greater than 30, normal approximation of the distribution is used irrespective of the original right skewedness of the variable.

The likelihood of the random sample of 277 individuals will result in a sample mean waist size of at least 86.3 cm is given as:

\[\begin{array}{c} P\left( {\bar X \ge 86.3} \right) = 1 - P\left( {\bar X \le 86.3} \right)\\ = 1 - P\left( {\frac{{\bar X - E\left( {\bar X} \right)}}{{{\sigma _{\bar X}}}} \le \frac{{86.3 - E\left( {\bar X} \right)}}{{{\sigma _{\bar X}}}}} \right)\\ = 1 - P\left( {Z \le \frac{{86.3 - 85}}{{0.32}}} \right)\\ = 1 - P\left( {Z \le 1.44} \right) \end{array}\]

Using standard normal tables (A.3), the value of probability $P\left( {\bar X \ge 86.3} \right)$ is 0.9251.

Therefore,

\[\begin{array}{c} P\left( {\bar X \ge 71} \right) = 1 - P\left( {Z \le 3.125} \right)\\ = 1 - 0.9251\\ = 0.0749 \end{array}\]

Therefore, the likelihood that a random sample of 277 individuals will result in a sample mean $\bar X$waist size of at least 86.3 cm is 0.0749.

 
Step 4

c.

Referring back to (b), suppose now that the population mean waist size in 82 cm. Required to find what is the (approximate) probability that the sample mean will be at least 86.3 cm.

The population mean ($\mu$) waist size is given as 85 cm and that the population standard deviation ($\sigma$) is 15 cm, with sample size (n) 277.

Using part (b) the sample mean, $\bar X$ is 82 and sample standard deviation, ${\sigma _{\bar X}}$= 0.9.

As the sample size (277) is greater than 30, normal approximation of the distribution is used.

The likelihood of the random sample of 277 individuals will result in a sample mean waist size of at least 86.3 cm is given as:

\[\begin{array}{c} P\left( {\bar X \ge 86.3} \right) = 1 - P\left( {\bar X \le 86.3} \right)\\ = 1 - P\left( {\frac{{\bar X - E\left( {\bar X} \right)}}{{{\sigma _{\bar X}}}} \le \frac{{86.3 - E\left( {\bar X} \right)}}{{{\sigma _{\bar X}}}}} \right)\\ = 1 - P\left( {Z \le \frac{{86.3 - 82}}{{0.32}}} \right)\\ = 1 - P\left( {Z \le 4.77} \right) \end{array}\]

The value of probability $P\left( {Z \ge 4} \right)$ is approximately 1, as the probability, $P\left( {Z \le 4.77} \right) > P\left( {Z \ge 4} \right)$.

Therefore, $P\left( {Z \le 4.77} \right)$ is approximately 1.

Therefore,

\[\begin{array}{c} P\left( {\bar X \ge 86} \right) = 1 - P\left( {Z \le 4.77} \right)\\ \approx 1 - 1\\ \approx 0 \end{array}\]

Therefore, the likelihood that a random sample of 277 individuals will result in a sample mean $\bar X$ waist size of at least 86.3 cm is 0.

Hence, 82 cm is a reasonable value for $\mu$.