4.113 Credit/debit card market shares. The following table reports the U.S. credit/debit card industry’s market share data for 2015. A random sample of 100 credit/debit card users is to be questioned regarding their satisfaction with

their card company. For simplification, assume that each card user carries just one card and that the market share percentages are the percentages of all card customers that carry each brand.

Credit debit Card	Market Share %
Visa	59
MasterCard	26
Discover	2
American Express	13

Source:Based on Nilson Reportdata, June 2015.

a. Propose a procedure for randomly selecting the 100 card users.

b. For random samples of 100 card users, what is the expected number of customers who carry Visa? Discover?

c. What is the approximate probability that half or more of the sample of card users carry Visa? American Express?

d. Justify the use of the normal approximation to the binomial in answering the question in part c.

A random sample of 100 credit/debit card users is to be questioned regarding their satisfaction with their card company

a.

Here a simple random sample can be used to draw samples from the population.

Here population unit has unique card numbers. So need to generate 100 random numbers based on the card numbers given, then based on these random numbers we can select the samples from the population.

b.

Here the selected random sample is 100 i.e, n=100

The proportion of Visa is 0.59 i.e, p=0.59

So, the expected value of Visa in the sample is,

$$\begin{gathered} \eta p=100\times 0.59 \\ =59 \end{gathered}$$

So, the expected Visa is 59

The proportion of Discover is 0.02 i.e, p=0.02

So, the expected value of Discover in the sample is,

$$\begin{gathered} \eta p=100\times 0.02 \\ =2 \end{gathered}$$

So the expected Discover is 2

c.

Let X be the number of Visa cards in the samples.

For Visa cards,

$$\begin{gathered} \mu =\eta p \\ =100\times 0.59 \\ =59 \\ \sigma =\sqrt{np(1-p)} \\ =\sqrt{100\times 0.59\times (1-0.59)} \\ =4.91833 \end{gathered}$$

The probability that half or more sample card users carry Visa is

role="math" localid="1660284771650" $$\begin{gathered} P(x\ge 50)=1-P(X<50) \\ =1-P\left(\frac{X-59}{4.91833}>\frac{50-59}{4.91833}\right) \\ =1-P(Z<1.8299) \\ =1-(1-P(Z<1.8299) \\ =P(Z<1.8299) \\ =0.96637 \end{gathered}$$

So, the probability is 0.96637

Let X be the number of American Express card users in the sample.

For Visa card,

$$\begin{gathered} \mu =np \\ =100\times 0.13 \\ =13 \end{gathered}$$

The probability that half or more sample card users carry American Express is

$$\begin{gathered} P(x\ge 50)=1-P(X<50) \\ =1-P\left(\frac{X-59}{3.36303}>\frac{13-59}{3.36303}\right) \\ =1-P(Z<13.6781) \\ =1-(1-P(Z<13.6781) \\ =P(Z<13.6781) \\ =1 \end{gathered}$$

So, the probability is 1

d.

For normal approximation to the binomial distribution, the distribution should satisfy two properties.

$np\ge 5andn(1-p)\ge 5$

For Visa

role="math" localid="1660284996733" $$\begin{gathered} np=100\times 0.59 \\ =59 \\ >5 \\ n(1-p)=100\times 0.41 \\ =41 \\ >5 \end{gathered}$$

So here two properties are satisfied, therefore a normal approximation can be used.

For American Express

$$\begin{gathered} np=100\times 0.13 \\ =13 \\ >5 \\ n(1-p)=100\times 0.87 \\ =87 \\ >5 \end{gathered}$$

Here also both the properties are satisfied, so we can use a normal approximation.