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Hoochie [10]
2 years ago
7

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

he 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).
Mathematics
1 answer:
julia-pushkina [17]2 years ago
5 0

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.

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