Answer:
7
Step-by-step explanation:
This is a Poisson distribution problem with the formula;
P(x;μ) = (e^(-μ)) × (μ^(x))/x!
We are told that the grocer sells three of a certain article per week. Thus;
μ = 3
Now, we want to find out How many of these should he have in stock so that the chance of his running out within a week is less than 0.01.
This means;
P(X > k) < 0.01
This can be rewritten as;
P(X ≤ k) < 0.99
Let's try x = 8
P(8;3) = (e^(-3)) × (3^(8))/8!
P(8;3) = 0.008
But; P(X ≤ 8) = 1 - 0.008 = 0.992
This is more than 0.99 and thus is not the answer
Let's try x = 7
P(7;3) = (e^(-3)) × (3^(7))/7!
P(7;3) = 0.022
But; P(X ≤ 7) = 1 - 0.022 = 0.978
Thus is less than 0.99.
Thus, stock should be 7
