Question

The number of rookie cards in a variety pack of baseball cards is normally distributed with a mean of 3 cards and the standard deviation of 1 card. If you purchase 10 variety packs, approximately how many packs would have between 2-4 rookie cards?

Answer 1

`z = (2-3)/((1)/(√(10)))=-3.163`

`z = (4-3)/((1)/(√(10)))=3.163`

`P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)`

`P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)= 0.9992 -0.00078= 0.99842`

So then we will expect 9.98 packages between 2-4 rookie cards in the sample of 10

Explanation:

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

`X \sim N(3,1)`

Where
`\mu=3` and
`\sigma=1`

We select a sample size of n = 10 variety packs and we want to find this probability:

`P(2<\bar X<4)`

We can use the z score formula given by:

`z=(x-\mu)/((\sigma)/(√(n)))`

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

We can find the z score for 2 and 4 and we got:

`z = (2-3)/((1)/(√(10)))=-3.163`

`z = (4-3)/((1)/(√(10)))=3.163`

So we can find the probability with this difference

`P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)`

And using the normal standard distirbution or excel we got:

`P(-3.163<z<3.163)=P(z<3.163)-P(z<3.163)= 0.9992 -0.00078= 0.99842`

So then we will expect 9.98 packages between 2-4 rookie cards in the sample of 10