SOLUTIONS TO EXERCISES IN CHAPTER 2



Answers PS #1

2.3 (a)

[pic]

(b) The probability that, on a given Monday, either 2, or 3, or 4 students will be absent is

[pic].

(c) The probability that, on a given Monday, more than 3 students are absent is

[pic]

d) [pic][pic]

3.066 is the average number of students absent on Mondays after considering infinitely many Mondays.

(e) [pic] (this is one way of calculating it…)

[pic]

[pic] = 10.5776

[pic] [pic]

The alternative is to use the formula E(X-()2 = (0-3.066)2x0.005 + (1-3.066)2 x 0.025 + …+ ((7-3.066)2 x 0.001 = 1.1776

(f) [pic] (Use the rule 2.3.3c on page 20)

[pic] (Use the rule 2.3.5 on page 21)

2.5 (a) The probability density function for x is

[pic] [pic]. Since our case has a = 10, b = 20, [pic] [pic]

(b) The total area beneath the pdf for 10 ( x ( 20 is the area of a rectangle. Namely,

Area = [pic]

c) [pic]

2.6 (a) For 4-year schools

[pic]

(b) Let X = 0 if a male is selected

X = 1 if a female is selected

Y = 1 if 4-year student is chosen

Y = 2 if 2-year student is chosen

Y = 3 if a less than 2-year student is chosen

[pic]

[pic]

[pic]

(c) For [pic] we note from part (b) that [pic] Similarly, for [pic] we obtain

[pic]

Thus, the marginal probability function [pic] is given by

[pic]

For [pic] we have

[pic]

(d) The conditional probability function for Y given that X = 1 can be obtained using the result

[pic]

Thus,

[pic]

(e) If a randomly chosen student is male, the probability that he attends a 2-year college is

[pic]

(f) If a randomly chosen student is a male, the probability that he attends either a 2-year college or a 4-year college is

[pic]

[pic]

[pic]

(g) For gender and type of college institution to be statistically independent, we need

[pic] for all x and y.

For x = 0 and y = 1, we have

[pic]

While these two values are close, they are not identical. Hence, gender and type of institution are not independent.

2.7 The mutual fund has an annual rate of return, [pic].

For each one of these problems, sketch a diagram of the normal curve and SHADE in the area that corresponds to the question. Then use the normal probability table to determine the actual area.

(a) [pic]

(b) [pic][pic]

(c) Now, [pic]

[pic]

[pic]

From the modification, the probability that a 1-year return will be negative is increased from 6.2% to 8.2%, and the probability that a 1-year return will exceed 15% is increased from 10.56% to 27.43%. Since the chance of negative return has increased only slightly, and the chance of a return above 15% has increased considerably, I would advise the fund managers to make the portfolio change.

2.9 not on list, but still good practice. Draw a normal distribution for each one of these and shade in the appropriate area:

(a) [pic]

(b) [pic]

(c) [pic]

(d) [pic]

(e) [pic]

(f) [pic]

g) Computer software files xr2-9.xls, contain the instructions for computing these probabilities.

2.12 (a) [pic]

(b) [pic] (this is the area under the line above from X=0 to X=1/2. Shade this area in on the graph. To calculate the area, remember that the area of a triangle is (1/2)*base*height.)

c) [pic] (this is the area under the line above between X=1/4 and X=3/4. shade your graph…i can’t seem do shade in Word)

2.15 [pic] for x = 0,1.

(a) The mean of the discrete random variable, X, is

[pic]

The variance of X is

[pic]

[pic]

(b) [pic]

Given [pic] are independent, we have

[pic]

(c) [pic]

[pic]

2.16

| | |Y | |

| | |1 |3 |9 | |

| |2 |1/8 |1/24 |1/12 |1/4 |

|X |4 |1/4 |1/4 |0 |1/2 |

| |6 |1/8 |1/24 |1/12 |1/4 |

| | |1/2 |1/3 |1/6 | |

(a) The marginal probability density function of Y is h(y) where

h(1) = 1/2 h(3) = 1/3 h(9) = 1/6

(b) The conditional probability density function for y is given by the equation [pic]. Applying this rule, we can obtain [pic], which is given by

|y | [pic] |

|1 | [pic] = 1/2 |

|3 |[pic] = 1/6 |

|9 |[pic] = 1/3 |

Therefore, the conditional probability density function, [pic], is given by

[pic] = 1/2 [pic] = 1/6 [pic] = 1/3

(c) [pic]

where [pic]

[pic]

From first principles

[pic]

Alternatively, using results in Section 2.5 it is possible to show that

[pic]

This is an important result that will be used throughout the text. In terms of the current example we can show that E(XY) = 12

E(XY) = (2)(1)(1/8) + (2)(3)(1/24) + (2)(9)(1/12)

+ (4)(1)(1/4) + (4)(3)(1/4) + (4)(9)(0)

+ (6)(1)(1/8) + (6)(3)(1/24) + (6)(9)(1/12) = 288/24 = 12

and hence that

[pic]

(d) This is an example where X and Y are not independent, despite the fact that their covariance is zero. If X and Y are independent, then [pic]. To prove that this result does not hold let us consider two outcomes. First, if x = 2 and y = 3,

[pic]

Also, if x = 4 and y = 9

[pic]

2.23 (a) [pic]

[pic]

|X \ Y |0 |1 | |

|0 |0.208 |0.120 |0.328 |

|1 |0.312 |0.360 |0.672 |

| |0.52 |0.48 |1.0 |

(b) By summing the relevant values we obtain these marginal distributions

|x |[pic] | |y |[pic] |

|0 |0.328 | |0 |0.52 |

|1 |0.672 | |1 |0.48 |

(c) This function is given by [pic]

| y |[pic] |[pic] |

|0 |[pic] |0.6341 |

|1 |[pic] |0.3659 |

(d) [pic] [pic]

(e) [pic] [pic]

(f) [pic] this is the term (1)(1)(0.36) = 0.36

(This sum has four terms but three of the four involve zero values, leaving only one of the four terms being non-zeroOnly the terms where both x and y are equal to one needs to be considered in the above summation. Other terms are zero.)

[pic]

(g) Use the rules on page 31:

[pic]

[pic]

2.25 Let X represent the life length of the computer. The fraction of computers lasting for a given time is equal to the probability of one computer, selected at random, lasting for that given time. For each probability below, you should draw a picture of the normal distribution and shade in the area that corresponds to the question. Then, standardize the value and use the table to determine the probability (area).

(a) [pic] [pic]

(b) [pic] [pic]

(c) [pic] [pic]

(d) [pic]

[pic]

(e) We want X0 such that P(X < X0) = 0.05. Now, P(0 < Z < 1.645) = 0.45 (from tables). Hence, P(Z < (1.645) = 0.05. Thus, an appropriate X0 is defined by

[pic] or [pic]

2.26 NOTE: skip this one…

-----------------------

[pic]

x

20

2

10

1

[pic]

[pic]

f(x)

x

1

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