Of a particular major car manufacturer, 40% of all cars made offer power locks a
ID: 3208618 • Letter: O
Question
Of a particular major car manufacturer, 40% of all cars made offer power locks as a standard feature and 95% offer passenger side airbags as a standard feature. If 5 cars are selected at random, what is the probability that none of these cars offers power locks as a standard feature? If 10 cars are selected at random, what is the probability that exactly nine cars will offer power locks as a standard feature? If 15 cars are selected at random, what is the probability that at least ten will have passenger side airbags as a standard feature? If 20 cars are selected at random, what is the probability that between ten and fifteen (inclusively) will have passenger side airbags as a standard feature? If 25 cars are selected at random, what is the expected number of cars which have passenger side airbags? What is the standard deviation of the number of cars which have passenger side airbags?Explanation / Answer
Binomial 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
a.
P( X = 0 ) = ( 5 0 ) * ( 0.4^0) * ( 1 - 0.4 )^5
= 0.0778
b.
P( X = 9 ) = ( 10 9 ) * ( 0.4^9) * ( 1 - 0.4 )^1
= 0.0016
c.
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 ) * 0.95^9 * ( 1- 0.95 ) ^6 + ( 15 8 ) * 0.95^8 * ( 1- 0.95 ) ^7 + ( 15 7 ) * 0.95^7 * ( 1- 0.95 ) ^8 + ( 15 6 ) * 0.95^6 * ( 1- 0.95 ) ^9 + ( 15 5 ) * 0.95^5 * ( 1- 0.95 ) ^10 + ( 15 4 ) * 0.95^4 * ( 1- 0.95 ) ^11 + ( 15 3 ) * 0.95^3 * ( 1- 0.95 ) ^12 + ( 15 2 ) * 0.95^2 * ( 1- 0.95 ) ^13 + ( 15 1 ) * 0.95^1 * ( 1- 0.95 ) ^14 + ( 15 0 ) * 0.95^0 * ( 1- 0.95 ) ^15
= 0.0001
P( X > = 10 ) = 1 - P( X < 10) = 0.9999
d.
P( X = 10 ) = ( 20 10 ) * ( 0.95^10) * ( 1 - 0.95 )^10
= 0.0000000108
P( X = 11 ) = ( 20 11 ) * ( 0.95^11) * ( 1 - 0.95 )^9
= 0.0000001866
P( X = 12 ) = ( 20 12 ) * ( 0.95^12) * ( 1 - 0.95 )^8
= 0.000002659
P( X = 13 ) = ( 20 13 ) * ( 0.95^13) * ( 1 - 0.95 )^7
= 0.0000310893
P( X = 14 ) = ( 20 14 ) * ( 0.95^14) * ( 1 - 0.95 )^6
= 0.0002953482
P( X = 15 ) = ( 20 15 ) * ( 0.95^15) * ( 1 - 0.95 )^5
= 0.002244646
P( 10 <= X <=15 ) = 0.0000000108 + 0.0000001866 + 0.000002659 + 0.0000310893 + 0.0002953482 + 0.002244646 = 0.0025739399
e.
Mean ( np ) =25 * 0.95 = 23.75
Standard Deviation ( npq )= 25*0.95*0.05 = 1.0897
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