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Fynjy0 [20]
3 years ago
15

On average, a furniture store sells four card tables in a week. Assuming a Poisson distribution for the weekly sales, the probab

ility that the store will sell exactly seven card tables in a given week is most nearly Select one: a. 0.11 b. 0.075 c. 0.15 d. 0.060
Mathematics
1 answer:
vivado [14]3 years ago
8 0

Answer:

Assuming a Poisson distribution for the weekly sales, the probability that the store will sell exactly seven card tables in a given week is 0.060

Step-by-step explanation:

In order to calculate the probability that the store will sell exactly seven card tables in a given week we would have to calculate the following formula:

probability that the store will sell exactly seven card tables in a given week= e∧-λ*λ∧x/x!

According to the given data furniture store sells four card tables in a week, hence λ=4

Therefore, probability that the store will sell exactly seven card tables in a given week=e∧-4*4∧7/7!

probability that the store will sell exactly seven card tables in a given week=0.060

Assuming a Poisson distribution for the weekly sales, the probability that the store will sell exactly seven card tables in a given week is 0.060

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Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

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The information provided is:

P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99

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Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

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Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188

Compute the probability that a randomly selected person is tested negative  as follows:

P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})

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▹ Answer

<em>b = -11</em>

▹ Step-by-Step Explanation

b + 9 = -2

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