Menopause in Mexico. In the article "Age at Menopause in Puebla. Mexico" (Human Biology, Vol. 75, No, 2, Pp. 205-206), authors L. Sievert and S. Hautaniemi compared the age of menopause for different populations. Menopause, the last menstrual period, is a universal phenomenon among females. According to the article, the mean age of menopause, surgical or natural, in Puebla, Mexico is 44.8 years with a standard deviation of 5.87 years. Let$\overline{x}$ denote the mean age of menopause for a sample of females in Puebla, Mexico.
a. For samples of size 40, find the mean and standard deviation of $\overline{x}$. Interpret your results in words.
b. Repeat part (a) with $n=120$.

It is given that the mean age of menopause, surgical or natural, in Puebla, Mexico is 44.8 years with a standard deviation of 5.87 years.

We know that the sample mean of a sample is equal to the population mean irrespective of the sample size.

The population mean in this case is given as $\mu =44.8$years.

So when the sample includes $40$females in Puebla, Mexico then the mean age of menopause, surgical or natural will be the same as the population mean.

So the sample mean is${\mu }_{\overline{x}}=44.8$years.

We know that the sample standard deviation of a sample is equal to the standard deviation of the variable under consideration divided by the square root of the sample size.

It is given that the standard deviation of the mean age of menopause is $\sigma =5.87$years.

So when the sample size is $40$, then its standard deviation of the variable$\overline{x}$is given as

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{40}} \\ {\sigma }_{\overline{x}}=\frac{5.87}{\sqrt{40}} \\ {\sigma }_{\overline{x}}\approx 0.93 \end{gathered}$$

Thus, the standard deviation of the mean age of menopause for a sample size of $40$is ${\sigma }_{\overline{x}}=0.93$years.

We know that the sample mean of a sample is equal to the population mean irrespective of the sample size.

The population mean in this case is given as $\mu =44.8$years.

So when the sample includes $120$females in Puebla, Mexico then the mean age of menopause, surgical or natural will be the same as the population mean.

So the sample mean is ${\mu }_{\overline{x}}=44.8$years.

We know that the sample standard deviation of a sample is equal to the standard deviation of the variable under consideration divided by the square root of the sample size.

It is given that the standard deviation of the mean age of menopause is $\sigma =5.87$years.

So when the sample size is $120$, then its standard deviation of the variable $\overline{x}$is given as

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{120}} \\ {\sigma }_{\overline{x}}=\frac{5.87}{\sqrt{120}} \\ {\sigma }_{\overline{x}}\approx 0.54 \end{gathered}$$

Thus, the standard deviation of the mean age of menopause for a sample size of $120$is ${\sigma }_{\overline{x}}=0.54$ years.