Answer:
The probability that a search of 900 randomly selected vehicles will find more than 175 with undeclared goods.
P( x > 175) = 0.0256
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 900
The estimated proportion 'p' = 0.17
Mean of the binomial distribution
μ = n p = 900 × 0.17 = 153
Standard deviation of the binomial distribution
σ = [tex]\sqrt{npq} = \sqrt{900 X 0.17 X 0.83} = 11.26[/tex]
Step(ii):-
Let 'X' be the random variable of normal distribution
mean μ= 153 and σ = 11.26
Given X = 175
[tex]Z = \frac{175-153}{11.268} = 1.95[/tex]
The probability that a search of 900 randomly selected vehicles will find more than 175 with undeclared goods.
P( x > 175) = P( Z> 1.95)
= 1- P( Z < 1.95)
= 1 - ( 0.5 +A( 1.95))
= 0.5 - A( 1.95)
= 0.5 - 0.4744
= 0.0256
Conclusion:-
The probability that a search of 900 randomly selected vehicles will find more than 175 with undeclared goods.
P( x > 175) = 0.0256
