91Ó°ÊÓ

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,

$$\begin{gathered} \mathrm{η}p=100×0.59 \\ =59 \end{gathered}$$

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,

$$\begin{gathered} \mathrm{η}p=100×0.02 \\ =2 \end{gathered}$$

So the expected Discover is 2

c.

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

For Visa cards,

$$\begin{gathered} \mathrm{μ}=\mathrm{η}p \\ =100×0.59 \\ =59 \\ \mathrm{σ}=\sqrt{np(1-p)} \\ =\sqrt{100×0.59×(1-0.59)} \\ =4.91833 \end{gathered}$$

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

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

So, the probability is 0.96637

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

For Visa card,

$$\begin{gathered} \mathrm{μ}=np \\ =100×0.13 \\ =13 \end{gathered}$$

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

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

So, the probability is 1

d.

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

$npâ‰\yen 5andn(1-p)â‰\yen 5$

For Visa

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

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

For American Express

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

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