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

Answer:

$z = \frac{2-3}{\frac{1}{\sqrt{10}}}=-3.163$

$z = \frac{4-3}{\frac{1}{\sqrt{10}}}=3.163$

$P(-3.163$

$P(-3.163$

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

Step-by-step 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:

$P(2$

We can use the z score formula given by:

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

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

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

$z = \frac{2-3}{\frac{1}{\sqrt{10}}}=-3.163$

$z = \frac{4-3}{\frac{1}{\sqrt{10}}}=3.163$

So we can find the probability with this difference

$P(-3.163$

And using the normal standard distirbution or excel we got:

$P(-3.163$

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