Historically, 12 percent of a mail-order firm\'s repeat charge-account customers

Question

Historically, 12 percent of a mail-order firm's repeat charge-account customers have an incorrect current address in the firm's computer database. The number of customers out of 19 who have an incorrect address in the database is a binomial random variable with n = 19 and = 0.12. (a) What is the probability that none of the next 19 repeat customers who call will have an incorrect address? (Round your answer to 4 decimal places.) (b) What is the probability that eight customers who call will have an incorrect address? (Round your answer to 4 decimal places.) Probability (c) What is the probability that nine customers who call will have an incorrect address? (Round your answer to 4 decimal places.) Probability (d) What is the probability that fewer than ten customers who call will have an incorrect address? (Round your answer to 4 decimal places.) Probability

Explanation / Answer

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 19 * 0.12
= 2.28
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 19 * 0.12 * 0.88
= 2.0064
III.
standard deviation = sqrt( variance ) = sqrt(2.0064)
=1.4165
a.
P( X = 0 ) = ( 19 0 ) * ( 0.12^0) * ( 1 - 0.12 )^19
= 0.0881
b.
P( X = 8 ) = ( 19 8 ) * ( 0.12^8) * ( 1 - 0.12 )^11
= 0.0008
c.
P( X = 9 ) = ( 19 9 ) * ( 0.12^9) * ( 1 - 0.12 )^10
= 0.0001
d.
P( X < 10) = P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 19 9 ) * 0.12^9 * ( 1- 0.12 ) ^10 + ( 19 8 ) * 0.12^8 * ( 1- 0.12 ) ^11 + ( 19 7 ) * 0.12^7 * ( 1- 0.12 ) ^12 + ( 19 6 ) * 0.12^6 * ( 1- 0.12 ) ^13 + ( 19 5 ) * 0.12^5 * ( 1- 0.12 ) ^14 + ( 19 4 ) * 0.12^4 * ( 1- 0.12 ) ^15 + ( 19 3 ) * 0.12^3 * ( 1- 0.12 ) ^16 + ( 19 2 ) * 0.12^2 * ( 1- 0.12 ) ^17 + ( 19 1 ) * 0.12^1 * ( 1- 0.12 ) ^18 + ( 19 0 ) * 0.12^0 * ( 1- 0.12 ) ^19
= 1

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.