Question

A statistics practitioner determined that the mean and stan-dard deviation of a data set were 90 and 15, respectively. What can yousay about the proportions of observations that lie between each of thefollowing intervals if you know that the distribution is bell-shaped? Whatcan you say if the distribution is not bell-shaped?

Answer 1

If the distribution is bell shaped we can approximate the probability with high accuracy using the z score formula.

a)
`P(75<X<105)`

And for this case we can use the z score given by:

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

And if we use it we got:

`P(75<X<105) =P((75-90)/(15) <Z<(105-90)/(15)) = P(-1< Z<1)= P(Z<1)-P(Z<-1) = 0.841-0.159=0.683`

b)
`P(60<X<120)`

And for this case we can use the z score given by:

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

And if we use it we got:

`P(60<X<120) =P((60-90)/(15) <Z<(120-90)/(15)) = P(-2< Z<2)= P(Z<2)-P(Z<-2) = 0.977-0.0228=0.955`

c)
`P(45<X<135)`

And for this case we can use the z score given by:

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

And if we use it we got:

`P(45<X<135) =P((45-90)/(15) <Z<(135-90)/(15)) = P(-3< Z<3)= P(Z<3)-P(Z<-3) = 0.999-0.0014=0.997`

If the distribution is NOT bell shaped the approximation with the z score NOT works and we need to have the distribution for X in order to find the probabilities.

Explanation:

Previous concepts

If the distribution is bell shaped we can approximate the probability with high accuracy using the z score formula.

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 of interest

We assume for this case that
`\mu=90` and
`\sigma=15`

We are interested on this probability

`P(75<X<105)`

And for this case we can use the z score given by:

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

And if we use it we got:

`P(75<X<105) =P((75-90)/(15) <Z<(105-90)/(15)) = P(-1< Z<1)= P(Z<1)-P(Z<-1) = 0.841-0.159=0.683`

Part b

`P(60<X<120)`

And for this case we can use the z score given by:

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

And if we use it we got:

`P(60<X<120) =P((60-90)/(15) <Z<(120-90)/(15)) = P(-2< Z<2)= P(Z<2)-P(Z<-2) = 0.977-0.0228=0.955`

Part c

`P(45<X<135)`

And for this case we can use the z score given by:

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

And if we use it we got:

`P(45<X<135) =P((45-90)/(15) <Z<(135-90)/(15)) = P(-3< Z<3)= P(Z<3)-P(Z<-3) = 0.999-0.0014=0.997`

If the distribution is NOT bell shaped the approximation with the z score NOT works and we need to have the distribution for X in order to find the probabilities.