On the average, a grocer sells three of a certain article per week. 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

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

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4