Statistics for Business & Economics 14e Metric Version Chapter5 - Chapter 5 Discrete Probability - Studocu

Chapter 5
Discrete Probability Distributions

Learning Objectives

Understand the concepts of a random variable and a probability distribution.
Be able to distinguish between discrete and continuous random variables.
Be able to compute and interpret the expected value, variance, and standard deviation for
a discrete random variable.
Be able to construct an empirical discrete distribution from available data.
Be able to compute the covariance and correlation coefficient for a bivariate empirical
discrete distribution.
Be able to compute and work with probabilities involving a binomial probability
distribution.
Be able to compute and work with probabilities involving a Poisson probability
distribution.
Know when and how to use the hypergeometric probability distribution.

Solutions:
1. a. Head, Head (H,H)
Head, Tail (H,T)
Tail, Head (T,H)
Tail, Tail (T,T)
b. x = number of heads on two coin tosses
c.

Outcome Values of x
(H,H) 2
(H,T) 1
(T,H) 1
(T,T) 0
d. Discrete. It may assume three values: 0, 1, and 2.
2. a. Let x = time (in minutes) to assemble the product.
b. It may assume any positive value: x > 0.
c. Continuous
3. Let Y = position is offered
N = position is not offered
a. S = {( Y,Y,Y ), ( Y,Y,N ), ( Y,N,Y ), ( N,Y,Y ), ( Y,N,N ), ( N,Y,N ), ( N,N,Y ), ( N,N,N )}
b. Let N = number of offers made; N is a discrete random variable.
c.
Experimental Outcome ( Y,Y,Y ) ( Y,Y,N ) ( Y,N,Y ) ( N,Y,Y ) ( Y,N,N ) ( N,Y,N ) ( N,N,Y ) ( N,N,N )
Value of N 3 2 2 2 1 1 1 0
4. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
5. a. S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
b.
Experimental Outcome (1,1) (1,2) (1,3) (2,1) (2,2) (2,3)
Number of Steps Required 2 3 4 3 4 5

Total 1.
b.

c. f(x ) ≥ 0 for x = 1,2,3,4.
Σ f(x ) = 1
9. a.

x f ( x )
1 ଶ଴଴଴ହ଴଺ ൌ 0.
2 ଶ଴଴଴ଷଽ଴ ൌ 0.
3 ଶ଴଴଴ଷଵ଴ ൌ 0.
4 ଶ଴଴଴ଶଵ଼ ൌ 0.
5 ଶ଴଴଴ହ଻଺ ൌ 0.
b. Each probability is ≥ 0 and .253 + .195 + .155 + .109 + .288 = 1.
c. f (4) + f(5) = .109 + .288 =.

f ( x )

x
1234

10. a.

x f ( x )
1 0.
2 0.
3 0.
4 0.
5 0.
1.
b.
x f ( x )
1 0.
2 0.
3 0.
4 0.
5 0.
1.
c. P (4 or 5) = f( 4) + f( 5) = 0 + 0 = 0.
d. Probability of very satisfied: 0.
e. Senior executives appear to be more satisfied than middle managers; 83% of senior
executives have a score of 4 or 5, with 41% reporting a 5. Only 28% of middle

b. f( 2) = 2/6 =.
c. f( 2) + f( 3) = 2/6 + 3/6 =.
14. a. f( 200) = 1 – f( –100) – f( 0) – f( 50) – f( 100) – f( 150)
= 1 – .95 =.
This is the probability MRA will have a $200,000 profit.
b. P (Profit) = f( 50) + f( 100) + f( 150) + f( 200)
= .30 + .25 + .10 + .05 =.
c. P (at least 100) = f( 100) + f( 150) + f( 200)
= .25 + .10 +.05 =.
15. a.

x f ( x ) xf ( x )
3 .25.
6 .50 3.
9 .25 2.
1 6.

Mục Lục

E ( x ) =  = 6

b.
x x – μ ( x – μ ) 2 f ( x ) ( x – μ ) 2 f ( x )
3 –3 9 .25 2.
6 0 0 .50 0.
9 3 9 .25 2.

4.

Var ( x ) =  2 = 4.

c.  = 4 = 2.

16. a.

y f ( y ) yf ( y )
2 .2.
4 .3 1.
7 .4 2.
8 .1.
1 5.

E ( y ) =  = 5.

b.
y y – μ ( y – μ ) 2 f ( y ) ( y – μ ) 2 f ( y )
2 –3 10 .20 2.
4 –1 1 .30.
7 1 3 .40 1.
8 2 7 .10.
4.
() 4.
4 2.

Var y





18. a/b/.

x f ( x ) xf ( x ) x – μ ( x – μ ) 2 ( x – μ ) 2 f ( x )
0 .2188 .0000 –1 1.
1 .5484 .5484 –.1825 .0333.
2 .1241 .2483 .8175 .6684.
3 .0489 .1466 1 3.
4 .0598 .2393 2 7.
Total 1 1 1.
 
E ( x ) Var ( x )
c/d.
y f ( y ) yf ( y ) y – μ ( y – μ ) 2 ( y – μ ) 2 f ( y )
0 .2497 .0000 –1 1.
1 .4816 .4816 –.2180 .0475.
2 .1401 .2801 .7820 .6115.
3 .0583 .1749 1 3.
4 .0703 .2814 2 7.
Total 1 1 1.
 

E ( y ) Var ( y )
e. The expected number of times that owner-occupied units have a water supply stoppage
lasting six or more hours in the past three months is 1, slightly less than the
expected value of 1 for renter-occupied units. And the variability is somewhat less
for owner-occupied units (1) as compared to renter-occupied units (1).
19. a. f(x ) ≥ 0 for all values of x.
Σ f(x ) = 1
Therefore, it is a valid probability distribution.
b. Probability x ≥ 40 is f( 40) + f( 45) + f( 50) = .15 + .10 + .10 =.
c. Probability x < 35 = f( 20) + f( 25) + f( 30) = .05 + .20 + .25 =.
d. Expected value, variance, and standard deviation computations follow.
x f ( x ) xf ( x ) x – μ ( x – μ ) 2 ( x – μ ) 2 f ( x )
20 .05 1 –14 203 10.
25 .20 5 –9 85 17.
30 .25 7 –4 18 4.
35 .15 5 0 0 0.
40 .15 6 5 33 4.
45 .10 4 10 115 11.
50 .10 5 15 248 24.
Total 1 34 73.
 

21. a. E ( x ) = Σ x f(x ) = 0(1) + 0(2) + 0(3) + 0(4) + 0(5) = 4.
b. E ( x ) = Σ x f(x ) = 0(1) + 0(2) + 0(3) + 0(4) + 0(5) = 3.

c. Executives:  2 = Σ( x – μ ) 2 f ( x ) = 1.

Middle Managers:  2 = Σ( x – μ ) 2 f ( x ) = 1.

d. Executives:  = 1.

Middle Managers:  = 1.

e. The senior executives have a higher average score: 4 versus 3 for the middle
managers. The executives also have a slightly higher standard deviation.
22. a. E ( x ) = Σ x f(x ) = 300 (.20) + 400 (.30) + 500 (.35) + 600 (.15) = 445
The monthly order quantity should be 445 units.
b. Cost: 445 @ $50 = $22,
Revenue: 300 @ $70 = 21,
$1,250 loss
23. a, b, and c follow.
The total number of responses is 1014, so f (0) = 365/1014 = .3600; f (1) = 264/1014 =
.2604, and so on.
x f(x ) xf(x ) x – μ ( x – μ ) 2 ( x – μ ) 2 f(x )
0 0 0 –1 1 0.
1 0 0 –0 0 0.
2 0 0 0 0 0.
3 0 0 1 2 0.

4 0 0 2 7 0.

Total 1 1.
1.
E ( x ) = 1 and Var ( x ) = 1.
d. The possible values of y are 1, 2, 3, and 4. The total number of responses is 649, so f (1)
= 264/649 = .41, f (2) = 193/649 = .30m and so on.
y f ( y ) yf ( y )
1 .4068.
2 .2974.
3 .1402.
4 .1556.
Total 1 2.
E ( y ) = 2. The expected value or mean number of cups per day for adults that drink
at least one cup of coffee on an average day is 2, or approximately a mean of two
cups per day. As expected, the mean is somewhat higher when we only take into
account adults who drink at least one cup of coffee per day.
24. a. Medium E ( x ) = Σ x f(x )
= 50 (.20) + 150 (.50) + 200 (.30) = 145
Large: E ( x ) = Σ x f(x )
= 0 (.20) + 100 (.50) + 300 (.30) = 140
Medium preferred.
b. Medium

80.

100.

x + y f ( x + y ) ( x + y ) f (x +
y )

x + y – E ( x + y ) [() x   yExy ] 2 () x   yExy fxy 2 

130 .2 26 34 1,156 231.

80 .5 40 –16 256 128.

100 .3 30 4 16 4.

E(x + y) = 96 Var(x + y) = 364

d.  xy [ ( ) Var x y   Var x ( ) Var y ( )]/ 2 (364 61 129) / 2 87 
Var(x) = 61 and Var(y) =129 were computed in part a, so
 x 61 7  y 129 11.

xy xy (7)(11) 87.
xy
 
 

The random variables x and y are positively related. Both the covariance and correlation
coefficient are positive. Indeed, they are very highly correlated; the correlation
coefficient is almost equal to 1.
e. Var x ( )      y Var x ( ) Var y ( ) 2 xy 61 129 2(87) 364
Var ( x ) + Var ( y ) = 61 + 129 = 190
The variance of the sum of x and y is greater than the sum of the variances by two times
the covariance: 2(87) = 174. It is positive because the variables are positively related in

this case. Whenever two random variables are positively related, the variance of the sum
of the randomly variables will be greater than the sum of the variances of the individual
random variables.
26. a. The standard deviation for these two stocks is the square root of the variance.

 x  Var x () 25 5%  y  Var y () 1 1%
Investments in stock 1 would be considered riskier than investments in stock 2 because
the standard deviation is higher. Note that if the return for stock 1 falls 8/5 = 1 or
more standard deviation below its expected value, an investor in that stock will
experience a loss. The return for stock 2 would have to fall 3 standard deviations
below its expected value before an investor in that stock would experience a loss.
b. Because x represents the percent return for investing in stock1, the expected return for
investing $100 in stock 1 is $8 and the standard deviation is $5. So to get the
expected return and standard deviation for a $500 investment, we just multiply by 5.
Expected return ($500 investment) = 5($8) = $42.
Standard deviation ($500 investment) = 5($5) = $25.
c. Because x represents the percent return for investing in stock 1 and y represents the
percent return for investing in stock 2, we want to compute the expected value and
variance for .5 x + .5 y.
E (.5 x + .5 y ) = .5 E ( x ) + .5 E ( y ) = .5(8) + .5(3) = 4 + 1 = 5.
22
22

(.5 .5 ) .5 ( ) .5 ( ) 2(.5)(.5)
(.5) (25) (.5) (1) 2(.5)(.5)( 3)
6 .25 1 5

Var x    y Var x Var y  xy



Meal Price ( y)
Quality ( x) 1 2 3 Total
1 0 0 0 0.
2 0 0 0 0.
3 0 0 0 0.
Total 0 0 0 1.
b. E(x) = 1(.28) + 2(.50) + 3(.22) = 1.
Var(x) = .28(1 – 1) 2 + .50(2– 1) 2 + .22(3 – 1) 2 =.
c. E(y) = 1(.26) + 2(.39) + 3(.35) = 2.
Var(y) = .26(1 – 2) 2 + .39(2 – 2) 2 + .35(3 – 2) 2 =.
d.  xy [Var( ) Var( ) Var( )]/ 2 [1 .4964 .6019]/ 2 .2854 xy     x y
Because the covariance  xy .2854is positive, we can conclude that as the quality rating
goes up, the meal price goes up. This is as we would expect.

.2854.
.4964.
 
xy  xyxy 
With a correlation coefficient of .5221 we would call this a moderately positive
relationship. It is not likely to find a low cost restaurant that is also high quality, but it is
possible. There are three of them lead to f (3,1) .01.

28. a. Marginal distribution of Direct Labor Cost

y f ( y ) y f ( y ) y – E ( y ) ( y – E ( y )) 2 ( y – E ( y )) 2 f ( y )
43 .3 12 –2 5 1.

b. Marginal distribution of Parts Cost

c. Let z = x + y represent total manufacturing cost (direct labor + parts).

45 .4 18 –.3 .09.
48 .3 14 2 7 2.
– 45.
– Var ( y ) = 3.
– E(y) = 45.
–  y = 1.
- 85 .45 38 –5 30 13. x f ( x ) xf ( x ) x – E ( x ) ( x – E ( x )) 2 ( x – E ( x )) 2 f ( x )
- 95 .55 52 4 20 11.
  – 90.
  – Var(x)= 24.
  – E(x) = 90.
  –  x= 4.
  - 1. z f ( z )