An automobile manufacturer is concerned about the possibility of a fault in the engine of a particular model of one of its cars. This fault could potentially cause the engine to fail to start. The average number of cars per year that experience this fault is 4 cars. If the probability that 10 or more cars experience this fault in a year is at least 0.01, the manufacturer will issue a recall. Will the manufacturer issue a recall for this engine fault

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.