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
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
Related searches
- chapter 2 developmental psychology quizlet
- medical terminology chapter 2 terms
- physics chapter 2 practice test
- psychology chapter 2 quizlet
- medical terminology chapter 2 test
- intro to psychology chapter 2 quiz
- introduction to psychology chapter 2 quizlet
- ncert solutions class 8 history chapter 1
- solutions to homelessness in california
- chapter 6 stock valuation solutions to questions and problems
- solutions to problems in education
- answers to the exerpt in chapter 2