Hi I have an exam coming up with these practise questions. can anyone help me answer them.
a book store/s records show that 40% of customers making a purchase used a credit card to make the payment. this morning 10 cutomers purchased books from the store.
a)find the probability that at least 2 cutomers used a credit card
b)find the probability that exactly 2 customers used a credit card
c)find the average number of customers who used a credit card
d)find the standard deviation of the number of customers who used a credit card.
a book store/s records show that 40% of customers making a purchase used a credit card to make the payment. this morning 10 cutomers purchased books from the store.
a)find the probability that at least 2 cutomers used a credit card
b)find the probability that exactly 2 customers used a credit card
c)find the average number of customers who used a credit card
d)find the standard deviation of the number of customers who used a credit card.
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I will need to assume that the 10 customers are independent. Then the number of customers (out of the 10) that use a credit card, X, is binomially distributed with n=10 trials and success probability p=0.4 .
So P(X = x) = (n choose x) p^x (1-p)^(n-x) = (10 choose x) (0.4)^x (1-0.4)^(10-x).
a) P(at least 2 customers use a credit card)
= P(X >= 2)
= 1 - P(X < 2)
= 1 - P(X = 0) - P(X = 1)
= 1 - (10 choose 0) (0.6)^10 - (10 choose 1) (0.4) (0.6)^9
= 1 - 1(0.6)^10 - 10(0.4)(0.6)^9
= 0.989922304 .
b) P(exactly 2 customers used a credit card)
= P(X = 2)
= (10 choose 2) (0.4)^2 (0.6)^8
= [10*9/(1*2)] (0.4)^2 (0.6)^8
= 0.120932352 .
c) The average number of customers who used a credit card is
E(X) = np = 10(0.4) = 4.
d) The variance of the number of customers who used a credit card is
Var(X) = npq = np(1-p) = 10(0.4)(0.6) = 2.4 .
Since standard deviation is the square root of variance, the standard deviation of the number of customers who used a credit card is SD(X) = sqrt(2.4) or about 1.5492 .
Lord bless you today!
So P(X = x) = (n choose x) p^x (1-p)^(n-x) = (10 choose x) (0.4)^x (1-0.4)^(10-x).
a) P(at least 2 customers use a credit card)
= P(X >= 2)
= 1 - P(X < 2)
= 1 - P(X = 0) - P(X = 1)
= 1 - (10 choose 0) (0.6)^10 - (10 choose 1) (0.4) (0.6)^9
= 1 - 1(0.6)^10 - 10(0.4)(0.6)^9
= 0.989922304 .
b) P(exactly 2 customers used a credit card)
= P(X = 2)
= (10 choose 2) (0.4)^2 (0.6)^8
= [10*9/(1*2)] (0.4)^2 (0.6)^8
= 0.120932352 .
c) The average number of customers who used a credit card is
E(X) = np = 10(0.4) = 4.
d) The variance of the number of customers who used a credit card is
Var(X) = npq = np(1-p) = 10(0.4)(0.6) = 2.4 .
Since standard deviation is the square root of variance, the standard deviation of the number of customers who used a credit card is SD(X) = sqrt(2.4) or about 1.5492 .
Lord bless you today!