Question

For 800 respondents to an inquiry of their age, results show a mean of 42 years and a standard deviation of 5 years. How many respondents fall between  2 standard deviation from the mean?

Answer 1

`P(32<X<52)=P((42-\mu)/(\sigma)<(X-\mu)/(\sigma)<(52-\mu)/(\sigma))=P((32-42)/(5)<Z<(52-42)/(5))=P(-2<z<2)`

And we can find this probability with ths difference:

`P(-2<z<2)=P(z<2)-P(z<-2)`

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

`P(-2<z<2)=P(z<2)-P(z<-2)=0.978-0.02275=0.9545`

So then we expect:

x = 0.9545*800= 763.6 or approximately 764 respondent within 2 deviations from the mean

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".

Solution to the problem

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

`X \sim N(42,5)`

Where
`\mu=42` and
`\sigma=5`

We are interested on this probability

`P(42-2*5<X<42+2*5)`

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

`z=(x-\mu)/(\sigma)`

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

`P(32<X<52)=P((42-\mu)/(\sigma)<(X-\mu)/(\sigma)<(52-\mu)/(\sigma))=P((32-42)/(5)<Z<(52-42)/(5))=P(-2<z<2)`

And we can find this probability with ths difference:

`P(-2<z<2)=P(z<2)-P(z<-2)`

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

`P(-2<z<2)=P(z<2)-P(z<-2)=0.978-0.02275=0.9545`

So then we expect:

x = 0.9545*800= 763.6 or approximately 764 respondent within 2 deviations from the mean