Question

Adult heights within a population are approximately normally distributed due to genetic and environmental variance. Height is partly determined by the interaction of 423 genes with 697 variants. Researchers conducted a study in 2016. They found the mean height of all adult males born between 1980 and 1994 to be 178.4 centimeters (cm) The standard deviation was 7.59 centimeters.

(https://ourworldindata.org/human-height#height-is-normally-distributed)

A random sample of 40 men born between 1980 and 1994 is taken.
a. (3pt) List the given numerical information with the correct symbols.









b. (3pt) Check the conditions to use the normal distribution to compute a probability.











c. (4pt) Compute the probability of getting a sample of 40 men born between 1980 and 1994 with a mean height between 178 and 180 cm. Show the calculator function and values used to get your answer. Round your answer to 4 decimal places.

Answer 1

$a. n=40, \mu=178.4, \sigma=7.59.$

b. We use the consequence of the Central Limit Theorem:

Suppose that the population from which samples are taken has a probability

distribution with mean $\mu$ and variance $\sigma^2$ that is not necessarily a normal

distribution. Then the standardised variable associated with $\overline{X}$, given by $Z=\frac{\overline{X}-\mu}{\sigma\sqrt{n}}$ is asymptotically normal, i. e. $\lim_{n\rightarrow\infty} P\{Z\leq z\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^ze^{-\frac{t^2}{2}}dt.$

Here $n=40>30$. Sample size is rather big and we can use this consequence. $c. P\{178\leq\overline{X}\leq180\}=F(180)-F(178)=\Phi(\frac{180-178.4}{(7.59)\sqrt{40}})-\Phi(\frac{178-178.4}{(7.59)\sqrt{40}})=0.0166\text{ where } F(x)=\frac{1}{\sqrt{2\pi}\sqrt{40}(7.59)}\int_{-\infty}^{x}e^{-\frac{(t-178.4)^2}{2\cdot 40\cdot(7.59)^2}}dt, \\ \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{t^2}{2}}dt.$