Chapter 05 - Discrete Random Variables
CHAPTER 5—Discrete Random Variables
5.1A random variable is a variable that assumes numerical values determined by the outcome of an experiment.
5.2The values of a discrete random variable can be counted, or listed; the values of a continuous random variable cannot be counted, or listed.
5.3a.Discrete
b.Discrete
c.Continuous
d.Discrete
e.Discrete
f.Continuous
g.Continuous
5.4See page 160 in text.
5.5The probability of each possible outcome is 0, and the sum of the probabilities of all possible outcomes is 1.
5.6The mean is the sum of each possible value of x multiplied by the probability of that value of x. This represents the mean expected value of x.
The standard deviation is the square root of the probability weighted sum of the squared deviations from the mean.
5.7The standard deviation measures the spread of the population of the random variable.
5.8a.Valid
b.Not valid;
c.Not valid;
d.Valid
5.9a. = 0(.2) + 1(.8) = .8
= (0 – .8)2(.2) + (1 – .8)2(.8) = .16
=
b.
c.
Since the probabilities sum to 1.00, ux is the mean of all possible observed values of x.
5.10a.x012
p(x)
b.x0123
p(x)
c.x01234
p(x)
5.11a.
contains at least of the
observed values of x.
contains at least of the
observed values of x.
b.See the methods outlined in part a.
c.See the methods outlined in part (a).
5.12a.x12345
p(x)
b.p(x) 0 for each value of x.
c.
d.
5.13a.Graph not included in this manual.
b.
If numerous oil wells were dug, the average profit would be $500.
5.14a.
b.contains at least of the observed values of x.
=[600 2(189.74)] = [220.52, 979.48]
contains at least of the observed values of x.
=[600 3(189.74)] = [30.78, 1169.22]
c.Most risk: 3, least risk: 1
Investment 1:
Investment 2:
Investment 3:
d.Investment A:
Investment B:
Investment B carries the greater risk.
e.Investment 1:
Investment 2:
Investment 3:
Most risk: 3
Least risk: 1
The results are the same because all = $600.
5.15a.x$400–$49,600
p(x).995.005
b.
c.x(.995) + (x – 50,000)(.005) = 1,000
x – 250 = 1,000
x = $1,250
5.16a.
b.
c.
d.B
e.A, smaller standard deviation and coefficient of variation.
f.Explanations will vary.
5.17Be sure to account for the cost of the ticket.
5.18a.
b.
c.
d.No; higher than A
e.A + C, A + C; lower
f.A + C
5.19a.x12345
p(x).1099.0879.3077.2967.1978
b.
- n identical trials
- each trial a success or failure
- probability of success constant for each trial
- trials are independent
5.21Values of x are the different possible numbers of successes in n trials.
5.22If the probability of a event occurring based on an assumption is very small and the event is actually observed, the assumption is considered false.
5.23 MTB > pdf;
SUBC> binom 5 .3.
BINOMIAL WITH N = 5 P = 0.300000
KP( X = K)
0 0.1681
1 0.3601
2 0.3087
3 0.1323
4 0.0284
50.0024
MTB > cdf;
SUBC> binom 5 .3.
BINOMIAL WITH N = 5 P = 0.300000
KP( X LESS OR = K)
0 0.1681
1 0.5282
2 0.8369
3 0.9692
4 0.9976
5 1.0000
a.
x = 0, 1, 2, 3, 4, 5
b.p(0) = .1681, p(1) = .3601, p(2) = .3087, p(3) = .1323, p(4) = .0284, p(5) = .0024
c.P(x = 3) = .1323
d.P(x 3) = .9692
e.P(x < 3 ) = P(x 2) = .8369
f.P(x 4) = 1 – P(x 3) = 1 – .9692 = .0308
g.P(x > 2) = 1 – P(x 2) = 1 – .8369 = .1631
h.
= 1.05
i.
P(–.54939 x 3.54939) = P(x 3) = .9692
5.24a.
b.(1) P(x = 5) = .0102
(2) P(x 3) = 1 – P(x 2) = 1 – .7443 = .2557
(3) P(x 2) = .7443
(4) P(x 1) = 1 – P(x = 0) = 1 – .1176 = .8824
5.25a.
b.(1) P(x 13) = .4509
(2) P(x > 10) = 1 – P(x 10) = 1 – .0127 = .9873
(3) P(x 14) = 1 – P(x 13) = 1 – .4509 = .5491
(4) P(9 x 12) = .0019 + .0105 + .0428 + .1285 = .1837
(5) P(x 9) = .0022
c.No, if the claim is true, then P(x 9) is very small.
5.26a.P(x = 0) = .7857
b.P(x 1) = 1 – P(x = 0) = 1 – .7857 = .2143
c.P(x 3) = .7857 + .1905 + .0221 + .0016 = .9999
d.P(x 2) = 1 – P(x 1) = 1 – .9762 = .0238
e.No, the probability of this result is small if the claim is true.
5.27a.
Compute the pdf and cdf in MINITAB using n = 4 and p = .5.
(1) P(x = 0) = .0625
(2) P(x > 2) = 1 – P(x 2) = 1 – .6875 = .3125
b.Compute the pdf and cdf in MINITAB using n = 20 and p = .50
(1) P(x 9) = .4119
(2) P(x > 11) = 1 – P(x 11) = 1 – .7483 = .2517
(3) P(x < 5) = .0059
c.No, if the claim is true, then the probability of fewer than 5 is very small (.0059).
5.28a.
P(x 4) = .9891
b.
P(x 12) = .9444
c.
P(x 1) = .9761
5.29a.Binomial with n = 25,
P(x = 0) = .9996
P(x 1) = 1 – P(x = 0) = .0004
b.Binomial with n = 25,
P(x = 0) = .4845
P(x 1) = 1 – .4845 = .5155
c.
d.
q = (.999)1/25
q = .999959981
p = 1 – .999959981 = .000040019
5.30The different possible number of times an event occurs in an interval of time or space.
- The probability of an event’s occurrence is the same for any two time intervals of equal length.
- Whether the event occurs in a time interval is independent of whether the event occurs in any other nonoverlapping interval.
5.32 MTB > pdf;
SUBC> pois 2.
POISSON WITH MEAN = 2.000
KP( X = K)
0 0.1353
1 0.2707
2 0.2707
3 0.1804
4 0.0902
5 0.0361
6 0.0120
7 0.0034
8 0.0009
9 0.0002
10 0.0000
MTB > cdf;
SUBC> pois 2.
POISSON WITH MEAN = 2.000
KP( X LESS OR = K)
0 0.1353
1 0.4060
2 0.6767
3 0.8571
4 0.9473
50.9834
6 0.9955
7 0.9989
8 0.9998
9 1.0000
a. for x = 0, 1, 2, 3, …
b. / x / 0 / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8p(x) / .1353 / .2707 / .2707 / .1804 / .0902 / .0361 / .0120 / .0034 / .0009
c.Graph not included in this manual.
d.P(x = 2) = .2707
e.P(x 4) = .9473
f.P(x < 4) = P(x 3) = .8571
g.P(x 1) = 1 – P(x = 0) = 1 – .1353 = .8647
P(x > 2) = 1 – P(x 2) = 1 – .6767 = .3233
h.P(1 x 4) = P(x 4) – P(x = 0) = .9473 – .1353 = .8120
i.P(2 < x < 5) = P(x = 3) + P(x = 4) = .1804 + .0902 = .2706
j.P(2 x < 6) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) = .5774
5.33a.
b.
P(–.828 x 4.828) = P(x4) = .9473
P(–2.424 x 6.242) = P(x6) = .9954
5.34Compute the pdf and cdf in MINITAB with = 7. Output not shown.
a.P(x = 10) = .0710
b.P(x 10) = .9015
c.P(x > 10) = 1 – P(x 10) = 1 – .9015 = .0985
5.35a.Compute the cdf in MINITAB with = 4; P(x 5) = .7851
b.Use the results from part (a); P(x > 5) = 1 – P(x 5) = 1 – .7851 = .2149
c.Compute the cdf in MINITAB with = 8; P(x 5) = .1912
d.Compute the cdf in MINITAB with = 6; P(x > 12) = 1 – P(x 12) = 1 – .9912 = .0088
5.36Compute the pdf in MINITAB with = 3.
a.P(x = 0) = .0498
b.Perhaps not; if the agency’s claim is true, the probability of no patrol cars is quite small.
c.
b.It is in control.
5.37Compute the cdf in MINITAB with = 1.8.
a.P(x 10) = 1 – P(x 9) = 1 – 1.000 = 0; Approximately zero
b.The hospital’s rate of comas is unusually high.
5.38 = np = 200(.005) = 1; compute the cdf in MINITAB with = 1.
a.P(x 4) = 1 – P(x 3) = 1 – .9810 = .0190
b.Probably not; the probability of 4 or more defectives is small if the claim is true.
5.39a.x012
p(x)
b.x012
p(x).16.48.36
c.x012
p(x).12.56.32
5.40a.x012
p(x)
b.x012
p(x).49.42.09
c.x012
p(x).54.42.04
5.41a.
b.
c.
d.4.39 b & c
e.4.39 b
5.42 for x = –2, –1, 0, 1, 2
a. / x / –2 / –1 / 0 / 1 / 2p(x) / / / / /
b.p(x) 0 for each value of x.
c.
d.
p(x) / .6 / .4
b.At least $6,000
5.44a.Graph not included in this manual.
b.
c.75%
d.
5.45a.Binomial, n = 8, p = .8
b.P(x 3) = .0104
c.No; if the claim is true, the probability of 3 or fewer being relieved is very small.
5.46a.Binomial, n = 5, p = .8
P(x 4) = 1 – P(x 3) = 1 – .2627 = .7373
b.Binomial, n = 25, p = .8
(1) P(x 15) = .0173
(2) P(x > 20) = 1 – P(x 20) = 1 – .5793 = .4207
(3) P(20 x 24) = P(x 24) – P(x 19) = .9962 – .3833 = .6129
(4) P(x = 24) = .0236
c. No; if the claim is true, the probability that 15 or fewer are not satisfied is only .0173.
5.47Poisson Distribution with = 3
a.
b.P(x 8) = .9962
c.1 – P(x 8) = .0038
d.Poisson with = 6
P(x 10) = .9574
e.Poisson with = 9
P(x 5) = .1157
5.48a.
b.P(x < 4) = P(x 3) = .9344
c.Poisson with = 6
P(x 12) = .9912
d.Poisson with = 6
P(x = 0) = .0025
5.49Binomial, n = 4, p = .85
P (x 1) = 1 – P(x <1) = 1 – .0005 = .9995
5.50Binomial, n = 20, p = .90
P(x 13) = .0024
Don’t believe the claim
5.51P(x 17) = .3232
Not much evidence against it.
5.52Poisson Distribution with = 8
P(x 17) = 1 – P(x <17) = 1 – P(x 16) = 1 – .9963 = .0037
Mean number of failures has increased.
5.53Poisson Distribution with = 10
P(x 4) = .0293
Claim is probably not true.
5.54P(x 8) = .3328
Not much evidence against the claim.
5-1