Question

Eighty-one percent of products come off the line within product specifications. Your quality control department selects 15 products randomly from the line each hour. Looking at the binomial distribution, if fewer than how many are within specifications would require that the production line be shut down (unusual) and repaired?

A) Fewer than 12

B) Fewer than 10

C) Fewer than 9

D) Fewer than 11

Answer 1

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 executed
p = success probability
mean = 15 * 0.81
= 12.15
II.
variance = npq
where
n = total number of repetitions experiment is executed
p = success probability
q = failure probability
variance = 15 * 0.81 * 0.19
= 2.3085
III.
standard deviation = sqrt( variance ) = sqrt(2.3085)
=1.5194

A.
P( X < 12) = P(X=11) + 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)    
= ( 15 11 ) * 0.81^11 * ( 1- 0.81 ) ^4 + ( 15 10 ) * 0.81^10 * ( 1- 0.81 ) ^5 + ( 15 9 ) * 0.81^9 * ( 1- 0.81 ) ^6 + ( 15 8 ) * 0.81^8 * ( 1- 0.81 ) ^7 + ( 15 7 ) * 0.81^7 * ( 1- 0.81 ) ^8 + ( 15 6 ) * 0.81^6 * ( 1- 0.81 ) ^9 + ( 15 5 ) * 0.81^5 * ( 1- 0.81 ) ^10 + ( 15 4 ) * 0.81^4 * ( 1- 0.81 ) ^11 + ( 15 3 ) * 0.81^3 * ( 1- 0.81 ) ^12 + ( 15 2 ) * 0.81^2 * ( 1- 0.81 ) ^13
= 0.3146
within specifications would require that the production line be repaired =0.3146*15 = 4.719= 5
B.
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)
= ( 15 9 ) * 0.81^9 * ( 1- 0.81 ) ^6 + ( 15 8 ) * 0.81^8 * ( 1- 0.81 ) ^7 + ( 15 7 ) * 0.81^7 * ( 1- 0.81 ) ^8 + ( 15 6 ) * 0.81^6 * ( 1- 0.81 ) ^9 + ( 15 5 ) * 0.81^5 * ( 1- 0.81 ) ^10 + ( 15 4 ) * 0.81^4 * ( 1- 0.81 ) ^11 + ( 15 3 ) * 0.81^3 * ( 1- 0.81 ) ^12 + ( 15 2 ) * 0.81^2 * ( 1- 0.81 ) ^13 + ( 15 1 ) * 0.81^1 * ( 1- 0.81 ) ^14 + ( 15 0 ) * 0.81^0 * ( 1- 0.81 ) ^15
= 0.049
within specifications would require that the production line be shut down (unusual) = 0.049*15 =0.735
C.
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)   
= ( 15 8 ) * 0.81^8 * ( 1- 0.81 ) ^7 + ( 15 7 ) * 0.81^7 * ( 1- 0.81 ) ^8 + ( 15 6 ) * 0.81^6 * ( 1- 0.81 ) ^9 + ( 15 5 ) * 0.81^5 * ( 1- 0.81 ) ^10 + ( 15 4 ) * 0.81^4 * ( 1- 0.81 ) ^11 + ( 15 3 ) * 0.81^3 * ( 1- 0.81 ) ^12 + ( 15 2 ) * 0.81^2 * ( 1- 0.81 ) ^13 + ( 15 1 ) * 0.81^1 * ( 1- 0.81 ) ^14 + ( 15 0 ) * 0.81^0 * ( 1- 0.81 ) ^15
= 0.0137
within specifications would require that the production line be shut down (unusual) = 0.0137*15 = 0.2055
D.
P( X < 11) = 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)    
= ( 15 10 ) * 0.81^10 * ( 1- 0.81 ) ^5 + ( 15 9 ) * 0.81^9 * ( 1- 0.81 ) ^6 + ( 15 8 ) * 0.81^8 * ( 1- 0.81 ) ^7 + ( 15 7 ) * 0.81^7 * ( 1- 0.81 ) ^8 + ( 15 6 ) * 0.81^6 * ( 1- 0.81 ) ^9 + ( 15 5 ) * 0.81^5 * ( 1- 0.81 ) ^10 + ( 15 4 ) * 0.81^4 * ( 1- 0.81 ) ^11 + ( 15 3 ) * 0.81^3 * ( 1- 0.81 ) ^12 + ( 15 2 ) * 0.81^2 * ( 1- 0.81 ) ^13 + ( 15 1 ) * 0.81^1 * ( 1- 0.81 ) ^14
= 0.1394
within specifications would require that the production line be repaired = 0.1394*15 = 2.091 =2