a)The probability that a customer selected will have more than two credit cards is 0.11 ,b)expected value is 1.34 and c)standard deviation is 0.96.
Given, probability distribution table
X p(X)
0 0.18
1 0.44
2 0.27
3 0.08
4 0.03
where,X= number of credit cards people own
- probability that a savings account customer selected at random will have more than two credit cards, p(X>2)=p(3)+P(4)=0.08+0.03=0.11
- Expected value of the number of credit cards a randomly selected customer owns Expected value or mean,[tex]\mu=\sum\limits^4_{n=0} X*P(X)=0*P(0)+1*P(1)+2*P(2)+3*P(3)+4*P(4)\\\\\mu=0+1(0.44)+2(0.27)+3(0.08)+4(0.03)\\\\\mu=0+0.44+0.54+0.24+0.12\\\\\mu=1.34[/tex]
- Standard deviation of number of credit cards. Standard deviation,[tex]\sigma=\sqrt{\sum(X-\mu)^2P(X)}\\\\=\sqrt{(-1.34)^2(0.18)+(0.34)^2(0.44)+(0.66)^2(0.27)+(1.66)^2(0.08)+(2.66)^2(0.03)}\\\\=\sqrt{(1.7956)(0.18)+(0.1156)(0.44)+(0.4356)(0.27)+(2.7556)(0.08)+(7.0756)(0.03)}\\\\=\sqrt{0.323208+0.050864+0.117612+0.220448+0.212268}\\\\=\sqrt{0.9244}\\\\=0.96[/tex]
Thus, the probability that a customer selected will have more than two credit cards is 0.11 ,expected value is 1.34 and standard deviation is 0.96.
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