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

Answer:

$P(32$

And we can find this probability with ths difference:

$P(-2$

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

$P(-2$

So then we expect:

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

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".  

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

$P(42-2*5$

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:

$P(32$

And we can find this probability with ths difference:

$P(-2$

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

$P(-2$

So then we expect:

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