Next 3 questions are related to the followinginformation The proportion of vehic

Question

Next 3 questions are related to the followinginformation The proportion of vehicles which drive above the speed limit on a freeway is 80%. Suppose 8 vehicles are randomly clocked. The following is the probability distribution of the number of vehicles which drive above the speed limit. (The blank cells are intentional!) f(x) 0.000003 1 0.000082 7 0.3355 0.1678 18 The probability that 6 vehicles drive above the speed limit is, 0.2936 0.2760 0.2594 0.2438 19 The probability that at most 5 vehicles drive above the speed limit is: 0.1228 0.2031 0.5033 0.7969

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
P( X = 0 ) = ( 8 0 ) * ( 0.8^0) * ( 1 - 0.8 )^8
= 0.0000026
P( X = 1 ) = ( 8 1 ) * ( 0.8^1) * ( 1 - 0.8 )^7
= 0.0000819
P( X = 2 ) = ( 8 2 ) * ( 0.8^2) * ( 1 - 0.8 )^6
= 0.0011469
P( X = 3 ) = ( 8 3 ) * ( 0.8^3) * ( 1 - 0.8 )^5
= 0.009175
P( X = 4 ) = ( 8 4 ) * ( 0.8^4) * ( 1 - 0.8 )^4
= 0.0458752
P( X = 5 ) = ( 8 5 ) * ( 0.8^5) * ( 1 - 0.8 )^3
= 0.1468006
P( X = 6 ) = ( 8 6 ) * ( 0.8^6) * ( 1 - 0.8 )^2
= 0.2936013
P( X = 7 ) = ( 8 7 ) * ( 0.8^7) * ( 1 - 0.8 )^1
= 0.3355443
P( X = 8 ) = ( 8 8 ) * ( 0.8^8) * ( 1 - 0.8 )^0
= 0.1677722
Q2.
P( X < = 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0) +   
= ( 8 5 ) * 0.8^5 * ( 1- 0.8 ) ^3 + ( 8 4 ) * 0.8^4 * ( 1- 0.8 ) ^4 + ( 8 3 ) * 0.8^3 * ( 1- 0.8 ) ^5 + ( 8 2 ) * 0.8^2 * ( 1- 0.8 ) ^6 + ( 8 1 ) * 0.8^1 * ( 1- 0.8 ) ^7 + ( 8 0 ) * 0.8^0 * ( 1- 0.8 ) ^8 +
= 0.2031
Q3.
Mean = np = 8 * 0.80 = 6.4
Mean amount collected is = 6.4 * 150 = 960

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