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Masteriza [31]
3 years ago
8

Insurance companies A and B each earn an annual profit that is normally distributed with the same positive mean. The standard de

viation of company A’s annual profit is one half of its mean. In a given year, the probability that company B has a loss (negative profit) is 0.9 times the probability that company A has a loss.Calculate the ratio of the standard deviation of company B’s annual profit to the standard deviation of company A’s annual profit.(A) 0.49(B) 0.90(C) 0.98(D) 1.11(E) 1.71
Mathematics
1 answer:
krok68 [10]3 years ago
7 0

Answer:

C. 0.98

Step-by-step explanation:

Let x be the mean of Company A and B annual profit and x/2 and y are standard deviation of Company A and B annual profit.

P(B<0) = 0.9*P(A<0)

P(Z<(0-x)/y) = 0.9*P(Z<(0-x)/(x/2))

P(Z<-x/y) = 0.9*P(Z<-2)

P(Z<-x/y) = 0.0205

x/y =2.04

Or y/x = 1 /2.05

y/x =0.49

Ratio of the standard deviation of company B annual profit to the standard deviation of company A annual profit =y/(x/2)

= 2*(y/x)

= 2*0.49

= 0.98

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Answer:

<em>Explanation to the answer below... (Answers in the Explanation</em><em>)</em>

Step-by-step explanation:

We Know that our rule is y = y = -6(3)^{x} , which means we can use this rule to find y from x. Since we don't know what X, is, we have to assume numbers like 1, to 7. And there is no space to write on the graph, I'll put a virtual copy of a graph.

1st Point,

y = -6(3)^{x}\\y =  -6(3)^{1}\\y = -6(3)\\y = -18\\x = 1, y = -18

2nd, point

y = -6(3)^x\\ y = -6(3)^2\\y = -6(9)\\y = -54\\x = 2, y = -54

3rd Point

y = -6(3)^x\\y = -6(3)^3\\y = -6(27)\\y = -162\\x = 3, y = -162

4th Point

y = -6(3)^4\\y = -6(81)\\y = -486\\x = 4, y = -486\\

5th Point

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6th Point

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Let's now plot them on the Graph, the values are so big, the graph might look a little odd.

Always remember that we can continue the points infinitely, and there is no stop, The Graph is attached below...

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Answer:

m∠BOC= 40 degrees

Step-by-step explanation:

A diagram has been drawn and attached below.

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Since ∠AOD is a straight line

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We are given that:

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From the diagram

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m∠BOC= 40 degrees

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