In: Math
. A bank has recently begun a new credit program. Customers meeting a certain credit requirement can obtain a credit card accepted by participating area merchants that carries a discount. Past data show that about 35% of all applicants for this card are rejected. If we let X represent the number of applicants who are approved for this credit card, the probabilities for the values of X can be calculated using the binomial probability distribution. Please answer the following questions based on the information given in this problem.
a. If 15 individuals apply for the credit card today what is the probability that none of them will be rejected? Show your work. (1 point)
b. If 15 individuals apply for the credit card today what is the probability of at most 5 of them being rejected? Show your work. (2 points)
c. If 15 individuals apply for the credit card today what is the probability of more than 12 of them being approved? Show your work. (2 points)
d. If 15 individuals apply for the credit card today what is the probability of less than half of them being approved? Show your work. (2 points)
e. If 15 individuals apply for the credit card today what is the probability of at least 6 of them being rejected? Show your work. (2 points)
f. How many rejections are expected out of the 15 applications? (1 point)
g. Calculate and interpret the standard deviation for the number of approvals out of the next 15 applications for the credit card. Show your work
p = 0.35
P(X = x) = nCx * px * (1 - p)n - x
a) P(X = 0) = 15C0 * (0.35)^0 * (0.65)^15 = 0.0016
b) P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= 15C0 * (0.35)^0 * (0.65)^15 + 15C1 * (0.35)^1 * (0.65)^14 + 15C2 * (0.35)^2 * (0.65)^13 + 15C3 * (0.35)^3 * (0.65)^12 + 15C4 * (0.35)^4 * (0.65)^11 + 15C5 * (0.35)^5 * (0.65)^10
= 0.5643
c) P(X > 12) = P(X = 13) + P(X = 14) + P(X = 15)
= 15C13 * (0.35)^13 * (0.65)^2 + 15C14 * (0.35)^14 * (0.65)^1 + 15C5 * (0.35)^15 * (0.65)^0
= 0.0001
d) P(X < 7.5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)
= 15C0 * (0.35)^0 * (0.65)^15 + 15C1 * (0.35)^1 * (0.65)^14 + 15C2 * (0.35)^2 * (0.65)^13 + 15C3 * (0.35)^3 * (0.65)^12 + 15C4 * (0.35)^4 * (0.65)^11 + 15C5 * (0.35)^5 * (0.65)^10 + 15C6 * (0.35)^6 * (0.65)^9 + 15C7 * (0.35)^7 * (0.65)^8
= 0.8868
e) P(X > 6) = 1 - P(X < 6)
= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))
= 1 - ( 15C0 * (0.35)^0 * (0.65)^15 + 15C1 * (0.35)^1 * (0.65)^14 + 15C2 * (0.35)^2 * (0.65)^13 + 15C3 * (0.35)^3 * (0.65)^12 + 15C4 * (0.35)^4 * (0.65)^11 + 15C5 * (0.35)^5 * (0.65)^10)
= 1 - 0.5643
= 0.4357
f) Mean = 15 * 0.35 = 5.25
g) Standard deviation = sqrt(np(1 - p))
= sqrt(15 * 0.35 * 0.65)
= 1.8473