Question

It is known from the medical records of millions of men from a large metropolis that the heights of grown men in this city follow a normal distribution with average height being 170.2 cm and 68% of men having a height between 162.6 cm and 177.8 cm. (a) If a random man is picked from this population, what is the probability that their height is more than 177.8 cm? (b) If a random sample of 9 men is picked, what is the probability that the mean height of men in this sample is more than 177.8 cm? What is the probability that the mean height of men in this sample is more than 174 cm? (c) Again, for a randomly chosen sample of 9 men, what is the probability that the mean height of the men in this sample is between 167.7 cm and 171.5 cm (d) If a large number of such random samples of 9 men are picked, what is the number such that 25% of these samples will have average height less than this number? (e) How does this number in the last part changes if the sample size is changed to 100 men in each sample? 

Answer 1

In statistics, the 68–95–99.7 rule, also known as the empirical rule, is a shorthand used to remember the percentage of values that lie within an interval estimate in a normal distribution: 68%, 95%, and 99.7% of the values lie within one, two, and three standard deviations of the mean, respectively.

In mathematical notation, these facts can be expressed as follows, where $X$  is an observation from a normally distributed random variable, $\mu$ is the mean of the distribution, and $\sigma$ is its standard deviation:

$$P(\mu-1\sigma\leq X\leq \mu+1\sigma)\approx68.27\%$$

$$P(\mu-2\sigma\leq X\leq \mu+2\sigma)\approx95.45\%$$

$$P(\mu-3\sigma\leq X\leq \mu+3\sigma)\approx99.73\%$$

a) Let $X=$ the height of the men: $X\sim N(\mu, \sigma^2)$

Given $\mu=170.2, P(162.6< X< 177.8)\approx68\%$

By the 68–95–99.7 rule,

$$P(\mu-1\sigma\leq X\leq \mu+1\sigma)\approx68.27\%$$

Then $\sigma=170.2-162.6=177.8-170.2=7.6.$

$$X\sim N(170.2, 7.6^2)$$

$$P(X>177.8)=1-P(X\leq177.8)$$

$$=1-P(Z\leq\dfrac{177.8-170.2}{7.6})=1-P(Z\leq1)$$

$$\approx0.158655$$

(b) Let $X=$ the mean height of men: $X\sim N(\mu, \sigma^2/n)$

Given $\mu=170.2, \sigma=7.6, n=9$

$$P(X>177.8)=1-P(X\leq177.8)$$

$$=1-P(Z\leq\dfrac{177.8-170.2}{7.6/\sqrt{9}})=1-P(Z\leq3)$$

$$\approx0.001350$$

$$P(X>174)=1-P(X\leq174)$$

$$=1-P(Z\leq\dfrac{174-170.2}{7.6/\sqrt{9}})=1-P(Z\leq1.5)$$

$$\approx0.066807$$

(c) Let $\bar{X}=$ the mean height of men: $\bar{X}\sim N(\mu, \sigma^2/n)$

Given $\mu=170.2, \sigma=7.6, n=9$

$$P(167.7<\bar{X}<171.5)$$

$$=P(\bar{X}<171.5)-P(\bar{X}\leq167.7)$$

$$=P(Z<\dfrac{171.5-170.2}{7.6/\sqrt{9}})-P(Z\leq\dfrac{167.7-170.2}{7.6/\sqrt{9}})$$

$$\approx P(Z<0.513158)-P(Z\leq-0.986842)$$

$$\approx0.696080-0.161860=0.534220$$

(d) Let $\bar{X}=$ the mean height of men: $\bar{X}\sim N(\mu, \sigma^2/n)$

Given $\mu=170.2, \sigma=7.6, n=9$

$$P(\bar{X}<x)=0.25$$

$$P(Z<z)=0.25$$

$$z\approx-0.674490$$

$$\dfrac{x-170.2}{7.6/\sqrt{9}}\approx-0.674490$$

$$x=168.5$$

(e) Let $\bar{X}=$ the mean height of men: $\bar{X}\sim N(\mu, \sigma^2/n)$

Given $\mu=170.2, \sigma=7.6, n=100$

$$P(\bar{X}<x)=0.25$$

$$P(Z<z)=0.25$$

$$z\approx-0.674490$$

$$\dfrac{x-170.2}{7.6/\sqrt{100}}\approx-0.674490$$

$$x=169.7$$

$$169.7>168.5$$

This number increases if the sample size is changed to 100 men in each sample.