Questions 1 to 7 refer to the following. Suppose that the price of a bond is uni

Question

Questions 1 to 7 refer to the following.

Suppose that the price of a bond is uniformly distributed between $80 and $85.

What is the expression for the probability density function of the bond price?

A) f(x) = 1/5 for 80 < x < 85, 0 elsewhere

B) f(x) = 1/80 for 80 < x < 85, 0 elsewhere

C) f(x) = 1/85 for 80 < x < 85, 0 elsewhere

D) f(x) = 1/165 for 80 < x < 85, 0 elsewhere

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Question 21 pts

What is the probability that the bond price will be more than $83?

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Question 31 pts

What is the probability that the bond price will be less than $84?

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Question 41 pts

What is the probability that the bond price will be equal to $85?

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Question 51 pts

What is the probability that the bond price will be between $81 and $85?

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Question 61 pts

Determine the expected price of the bond.

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Question 71 pts

Determine the standard deviation of the bond price.

Explanation / Answer

Given the price of a bond is uniformly distributed between 80 and 85. It can be divided into five equal parts as

80-81,81-82,82-83,83-84,84-85.

(a) So, the probability density function of the bond price is given as

f(x) = 1/5 for 80 < x < 85, 0 elsewhere

(b) P(X>83) = P(83<X<85). It includes 83-84 and 84-85

=>P(X>83) = P(83<X<85) = 2/5 = 0.4

(c) The range 80<X<84 covers 4 out of 5 equal parts.

=>P(X<84) = P(80<X<84) = 4/5 = 0.8

(d) P(X=85) = 0 because it is not included in the range as per the probability density function.

(e) P(81<X<85) includes 4 regions. So, P(81<X<85) = 4/5 = 0.8

(f) Expected value of bond = (80+85)/2 = 165/2 = 82.5 = E(x)

(g) Variance = E(X2)-E(X)2

E(X2) = (802+852)/2 = 6812.5

=> variance = 6812.5 - (82.5)2 = 6.25

=> standard deviation = Sqrt(variance) = sqrt(6.25) = 2.5. which should be option D.

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