Question

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with the mean 3, find the probability that the shop sells. . (a) At least 3 in a week. (b) At most 7 in a week. (c) More than 20 in a month (4 weeks).

Answer 1

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$, if $\lambda>10$.

Poisson variable with the mean 3

This means that $\lambda= 3$.

(a) At least 3 in a week.

This is $P(X \geq 3)$. So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

$\mu = \lambda = 4(3) = 12$

$\sigma = \sqrt{\lambda} = \sqrt{12}$

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 12}{\sqrt{12}}$

$Z = 2.31$

$Z = 2.31$ has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Answer 2

The probability of the selling the video recorders for considered cases are:

• P(At least 3 in a week) = 0.5768 approximately.
• P(At most 7 in a week) = 0.9881 approximately.
• P( more than 20 in a month) = 0.0839 approximately.

What are some of the properties of Poisson distribution?

Let X ~ Pois(λ)

Then we have:

E(X) = λ = Var(X)

Since standard deviation is square root (positive) of variance,

Thus,

Standard deviation of X = $\sqrt{\lambda}$

Its probability function is given by

f(k; λ) = Pr(X = k) = $\dfrac{\lambda^{k}e^{-\lambda}}{k!}$

For this case, let we have:

X = the number of weekly demand of video recorder for the considered shop.

Then, by the given data, we have:

X ~ Pois(λ=3)

Evaluating each event's probability:

Case 1: At least 3 in a week.

$P(X > 3) = 1- P(X \leq 2) = \sum_{i=0}^{2}P(X=i) = \sum_{i=0}^{2} \dfrac{3^ie^{-3}}{i!}\\\\P(X > 3) = 1 - e^{-3} \times \left( 1 + 3 + 9/2\right) \approx 1 - 0.4232 = 0.5768$

Case 2: At most 7 in a week.

$P(X \leq 7) = \sum_{i=0}^{7}P(X=i) = \sum_{i=0}^{7} \dfrac{3^ie^{-3}}{i!}\\\\P(X \leq 7) = e^{-3} \times \left( 1 + 3 + 9/2 + 27/6 + 81/24 + 243/120 + 729/720 + 2187/5040\right)\\\\P(X \leq 7) \approx 0.9881$

Case 3: More than 20 in a month(4 weeks)

That means more than 5 in a week on average.

$P(X > 5) = 1- P(X \leq 5) =\sum_{i=0}^{5}P(X=i) = \sum_{i=0}^{5} \dfrac{3^ie^{-3}}{i!}\\\\P(X > 5) = 1- e^{-3}( 1 + 3 + 9/2 + 27/6 + 81/24 + 243/120)\\\\P(X > 5) \approx 1 - 0.9161 \\ P(X > 5) \approx 0.0839$

Thus, the probability of the selling the video recorders for considered cases are:

• P(At least 3 in a week) = 0.5768 approximately.
• P(At most 7 in a week) = 0.9881 approximately.
• P( more than 20 in a month) = 0.0839 approximately.

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