Answer:
The probability of 10 or more cars experiencing the fault, 0.00813 is less than 0.01, hence, the manager will not issue a recall for this engine fault.
Step-by-step explanation:
This is Poisson distribution problem
Poisson distribution formula is given by
P(X = x) = (e^-λ)(λˣ)/x!
where λ = mean = 4 cars developing faults per year.
x = variable whose probability is required = 10 or more cars
Meaning we require
P(X ≥ x) = Σ (e^-λ)(λˣ)/x! (Summation From x to the end of the variables)
But Mathematically,
P(X ≥ 10) = 1 - P(X ≤ 9) (since only whole number values are allowed for X)
Therefore,
P(X ≥ 10) = 1 - P(X ≤ 9) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) +.......+ P(X=9)]
P(X ≥ 10) = 1 - P(X ≤ 9) = 1 - 0.99187 = 0.00813
P(X ≥ 10) = 0.00813 < 0.01, Hence the manager will not issue a recall for this engine fault.