Question

In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once.

a. In words, define the random variable X.

b. List the values that X may take on.

c. Give the distribution of X. X ~ _____(_____,_____)

d. How many pages do you expect to advertise footwear on them?

e. Calculate the standard deviation.

Short answer

a. The random variable X is the number of pages that advertise footwear.

b. The values of X is $0,1,2,3,...........,20$

c. The distribution of X is$X~Hypergeometric(20,\frac{29}{192})$

d. The number of pages to advertise footwear is $3.03$

e. The standard deviation is $1.5197$

Step-by-step solution

Content Introduction

A discrete probability distribution, the hypergeometric distribution. It's utilised when you want to know how likely it is to get a certain number of successes from a given sample size without replacement.

Explanation (part a)

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of pages that advertise footwear.

Explanation (part  b)

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand so,

$X=0,1,2,3,........20.$

Explanation (part c)

The random variable X refers to the number of trials and each trial is independent of others and has same probability of success. This implies that random variable X follows Hypergeometric Distribution

$X~Hypergeometric(20,\frac{29}{192})$

Explanation (part d)

The expected number pages to advertise footwear are

$E\left(X\right)=np$where, $n$is the random survey of pages and $p$is number of pages

$$\begin{gathered} E\left(X\right)=20\times \frac{29}{192} \\ E\left(X\right)=3.03 \end{gathered}$$

Explanation (part e)

The standard deviation is $\delta =\sqrt{np(1-p)}$

where, $n=20,p=\frac{29}{192}$

$$\begin{gathered} \delta =\sqrt{\left(20\right)\left(\frac{29}{192}\right)\times (1-\frac{29}{192})} \\ \delta =\sqrt{(3.02)\times (0.848)} \\ \delta =1.5197 \end{gathered}$$