Question

1. Data are from a normal distribution and the mean is 20 with a standard deviation of 2. a. What % of observations fall between 18 and 22

Answer 1

`P(18<X<22)=P((18-\mu)/(\sigma)<(X-\mu)/(\sigma)<(22-\mu)/(\sigma))=P((18-20)/(2)<Z<(22-20)/(2))=P(-1<z<1)`

And we can find this probability with the normal standard distribution and we got:

`P(-1<z<1)=P(z<1)-P(z<-1)=0.841 -0.159= 0.682`

Explanation:

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

`X \sim N(20,2)`

Where
`\mu=20` and
`\sigma=2`

We are interested on this probability

`P(18<X<22)`

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

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

Using this formula we got:

`P(18<X<22)=P((18-\mu)/(\sigma)<(X-\mu)/(\sigma)<(22-\mu)/(\sigma))=P((18-20)/(2)<Z<(22-20)/(2))=P(-1<z<1)`

And we can find this probability with the normal standard distribution and we got:

`P(-1<z<1)=P(z<1)-P(z<-1)=0.841 -0.159= 0.682`