Question

Height of Dutch Men. Males in the Netherlands are the tallest, on average, in the world with an average height of 183 centimeters (cm) (BBC News website). Assume that the height of men in the Netherlands is normally distributed with a mean of 183 cm and standard deviation of 10.5 cm.a.What is the probability that a Dutch male is shorter than 175 cm?b.What is the probability that a Dutch male is taller than 195 cm?c.What is the probability that a Dutch male is between 173 and 193 cm?d.Out of a random sample of 1000 Dutch men, how many would we expect to be taller than 190 cm?

Answer 1

Answer:

a) $P(X<175) = P(Z<\frac{175-183}{10.5}) =P(Z<-0.762)$

$P(z<-0.762)=0.223$

b) $P(X>195) =P(Z>\frac{195-183}{10.5})=P(Z>1.143)$

$P(z>1.143)=1-P(Z<1.143) =1-0.873=0.127$

c) $P(173<X<193)=P(\frac{173-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{193-\mu}{\sigma})=P(\frac{173-183}{10.5}<Z<\frac{193-183}{10.5})=P(-0.953<z<0.953)$

$P(-0.953<z<0.953)=P(z<0.953)-P(z<-0.953)$

$P(-0.953<z<0.953)=P(z<0.953)-P(z<-0.953)=0.830-0.170=0.660$

d) $P(\bar X >190)=P(Z>\frac{190-183}{\frac{10.5}{\sqrt{1000}}}=21.08)$

And using a calculator, excel ir the normal standard table we have that:

$P(Z>21.08)=1-P(Z<21.08) \approx 0$

So for this case we expect anyone with a heigth higher than 190 cm in a random sample of 1000.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(183,10.5)$  

Where $\mu=183$ and $\sigma=10.5$

We are interested on this probability

$P(X<175)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X<175) = P(Z<\frac{175-183}{10.5}) =P(Z<-0.762)$

And we can find this probability with the normal standard table or with excel:

$P(z<-0.762)=0.223$

Part b

$P(X>195)$

$P(X>195) =P(Z>\frac{195-183}{10.5})=P(Z>1.143)$

And we can find this probability with the normal standard table or with excel: using the complement rule

$P(z>1.143)=1-P(Z<1.143) =1-0.873=0.127$

Part c

$P(173<X<193)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(173<X<193)=P(\frac{173-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{193-\mu}{\sigma})=P(\frac{173-183}{10.5}<Z<\frac{193-183}{10.5})=P(-0.952<z<0.953)$And we can find this probability on this way:

$P(-0.953<z<0.953)=P(z<0.953)-P(z<-0.953)$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-0.953<z<0.953)=P(z<0.953)-P(z<-0.953)=0.830-0.170=0.660$Part d

Since the distributiion for X is normal then the distribution for the sample mean is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we eant this probability:

$P(\bar X >190)=P(Z>\frac{190-183}{\frac{10.5}{\sqrt{1000}}}=21.08)$

And using a calculator, excel ir the normal standard table we have that:

$P(Z>21.08)=1-P(Z<21.08) \approx 0$

So for this case we expect anyone with a heigth higher than 190 cm in a random sample of 1000.