Question

Assuming a normal distribution of 1000 cases, how many cases will be farther away from the mean than + 3 standard deviations?a.About 3 b.At least 500 c.It is impossible to estimate d.327

Answer 1

Answer:

$P(-3<Z<3)=P(Z<3)-P(Z<-3)=0.99865-0.00135=0.9973$

And then the probability that  will be farther away from the mean than + 3 standard deviations is just

1-0.9973= 0.0027

And the number of cases would be 1000*0.0027=2.7 or approximatley 3 cases. And the bes option is : a.About 3

Step-by-step explanation:

1) 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".  

2) 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:

$X \sim N(\mu,\sigma)$  

We are interested on this probability

$P(X-3\mu<X<X+3\mu)$

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

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

And we can find this probability on this way:

$P(-3<Z<3)=P(Z<3)-P(Z<-3)$

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

$P(-3<Z<3)=P(Z<3)-P(Z<-3)=0.99865-0.00135=0.9973$

And then the probability that  will be farther away from the mean than + 3 standard deviations is just

1-0.9973= 0.0027

And the number of cases would be 1000*0.0027=2.7 or approximatley 3 cases. And the bes option is : a.About 3