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olga_2 [115]
2 years ago
10

The editor of a textbook publishing company is deciding whether to publish a proposed textbook. Information on previous textbook

s published show that 20 % are huge​ successes, 30 % are modest​ successes, 30 % break​ even, and 20 % are losers. Before a decision is​ made, the book will be reviewed. In the​ past, 99 % of the huge successes received favorable​ reviews, 70 % of the moderate successes received favorable​ reviews, 40 % of the​ break-even books received favorable​ reviews, and 20 % of the losers received favorable reviews. If the textbook receives a favorable​ review, what is the probability that it will be huge​ success?
Mathematics
1 answer:
kogti [31]2 years ago
4 0

Answer:

34.86% probability that it will be huge​ success

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

P(B|A) = \frac{P(B)*P(A|B)}{P(A)}

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Receiving a favorable review.

Event B: Being a huge success.

Information on previous textbooks published show that 20 % are huge​ successes

This means that P(B) = 0.2

99 % of the huge successes received favorable​ reviews

This means that P(A|B) = 0.99

Probability of receiving a favorable review:

20% are huge​ successes. Of those, 99% receive favorable reviews.

30% are modest​ successes. Of those, 70% receive favorable reviews.

30% break​ even. Of those, 40% receive favorable reviews.

20% are losers. Of those, 20% receive favorable reviews.

Then

P(A) = 0.2*0.99 + 0.3*0.7 + 0.3*0.4 + 0.2*0.2 = 0.568

Finally

P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.2*0.99}{0.568} = 0.3486

34.86% probability that it will be huge​ success

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2 years ago
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Answer:

The answer is below

Step-by-step explanation:

A sugar refinery has three processing plants, all receiving raw sugar in bulk. The amount of raw sugar (in tons) that one plant can process in one day can be modelled using an exponential distribution with mean of 4 tons for each of three plants. If each plant operates independently,a.Find the probability that any given plant processes more than 5 tons of raw sugar on a given day.b.Find the probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day.c.How much raw sugar should be stocked for the plant each day so that the chance of running out of the raw sugar is only 0.05?

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f(x)=\lambda e^{-\lambda x}\\But\ \lambda= 1/\mu=1/4 = 0.25\\Therefore:\\f(x)=0.25e^{-0.25x}\\

a) P(x > 5) = \int\limits^\infty_5 {f(x)} \, dx =\int\limits^\infty_5 {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_5=e^{-1.25}=0.2865

b) Probability that exactly two of the three plants process more than 5 tons of raw sugar on a given day can be solved when considered as a binomial.

That is P(2 of the three plant use more than five tons) = C(3,2) × [P(x > 5)]² × (1-P(x > 5)) = 3(0.2865²)(1-0.2865) = 0.1757

c) Let b be the amount of raw sugar should be stocked for the plant each day.

P(x > a) = \int\limits^\infty_a {f(x)} \, dx =\int\limits^\infty_a {0.25e^{-0.25x}} \, dx =-e^{-0.25x}|^\infty_a=e^{-0.25a}

But P(x > a) = 0.05

Therefore:

e^{-0.25a}=0.05\\ln[e^{-0.25a}]=ln(0.05)\\-0.25a=-2.9957\\a=11.98

a  ≅ 12

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ollegr [7]

Answer:

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Step-by-step explanation:

Let

x ----> the number of dimes

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<em> Suri's coin purse</em>

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