Chapter 06 - Continuous Random Variables

Chapter 06 - Continuous Random Variables

CHAPTER 6—Continuous Random Variables

6.1Intervals of values.

6.2By finding areas under the curve.

6.3f(x)  0 for all x; area under the curve equals 1.

6.4Relative likelihood that x will be near the given point.

6.5When a variable has a rectangular distribution over a certain interval.

6.6MTB > cdf cl;

SUBC> uniform 2.0 8.0.

CONTINUOUS UNIFORM ON 2.0 TO 8.0

K P( X LESS OR = K)

0.00000.0000

0.50000.0000

1.00000.0000

1.50000.0000

2.00000.0000

2.50000.0833

3.00000.1667

3.50000.2500

4.00000.3333

4.50000.4167

5.00000.5000

5.50000.5833

6.00000.6667

6.50000.7500

7.00000.8333

7.50000.9167

8.00001.0000

a. for 2 x 8

= 0 otherwise

b.Graph not included in this manual.

c.P(3 x 5) = P(x 5) – P(x 3) = .5 – .1667 = .3333

d.P(1.5 x 6.5) = P(x 6.5) – P(x 1.5) = .75 – 0 = .75

e.

f.
P(1.536 x 8.464) = P(x 8) = 1

6.7

6.8a. for 0 x 6.

= 0 otherwise

b. f(x)

1/6

0 6 min

c.

d.

e.

6.9a.

b.

6.10

6.11a.for 120  x  140

b. f(x)

1/20

120 140 min

c. P(125  x  135) = 10 (1/20) = .5

d. P(x  135) = 5 (1/20) = .25

6.12a.

b.

6.13

6.14a. for 3 x 6.

b. f(x)

1/3

3 6 inches

c.P(x 4) = 1 – P(x < 4) = 1 – .3333 = .6667
P(x 5) = 1 – P(x < 5) = 1 – .6667 = .3333

6.15a.

b.

P(2.768 x 6.232) = P(3 x 6) = 1

6.16Entire family of normal distributions

Highest point on curve is mean, median and mode

Distribution is symmetrical

Tails go to infinity in both directions

Area right of  = Area left of  = 0.5

6.17a.center

b.spread

6.1868.26%, 95.44%, 99.73%

6.19Subtract the mean and divide the result by the standard deviation; tells us the number of standard deviations the value is above or below the mean.

6.20a.x equals the mean

b.x greater than the mean

c.x less than the mean

6.21The normal table provides the areas under the standard normal curve (the distribution of the z values).

6.22a.

6.23a.; x is one standard deviation below the mean.

b.; x is three standard deviations below the mean.

c.; x is equal to the mean.

d.; x is two standard deviations above the mean.

e.; x is four standard deviations above the mean.

6.24a.P(0 z 1.5) = 0.9332 – 0.5000 = .4332

b.P(z 2) = 1 – .9772 = .0228

c.P(z 1.5) =.9332

d.P(z –1) = 1 – .1587 =.8413

e.P(z –3) = .00135

f.P(–1 z 1) = .8413 – .1587 = .6826

g.P(–2.5 z .5) = .6915 – .0062 = .6853

h.P(1.5 z 2) = .9772 – .9332 = .0440

i.PP(–2 z –.5) = .3085 – .0228 = .2857

6.25a.

b.

c.

d.

e.

f.

6.26Restate each probability in terms of the standard normal random variable z.

Then use the table to solve.

a.P(1000  x  1200) = P(0  z  2) = .9772 – .5 = .4772

b.P(x > 1257) = P(z > 2.57) = 1 – .9949 = .005

c.P(x < 1035) = P(z < .35) = .6368

d.P(857  x  1183) = P(–1.43  z  1.83) = .9664 – .0764 = .8900

e.P(x  700) = P(z  – 3) = .00135

f.P(812  x  913) = P(–1.88  z  –.87) = .1922 – .0301 = .1621

g.P(x > 891) = P(z > –1.09) = 1 – .1379 = .8621

h.P(1050  x  1250) = P(.5  z  2.5) = .9938 – .6915 = .3023

6.27First find the z-value from the table that makes the statement true. Then calculate x using the formula:
x = z+ = z(100) + 500

a.P(x 696) = .025

b.P(x 664.5) = .05

c.P(x < 304) = .025

d.P(x 283) = .015

e.P(x < 717) = .985

f.P(x > 335.5) = .95

g.P(x 696) = .975

h.P(x 700) = .0228

i.P(x > 300) = .9772

6.28a.

b.

c.(1) P(x > 140) = P(z2.5) = 1 – .9938 = .0062
(2) P(x < 88) = P(z < –.75) = .2266
(3) P(72 < x < 128) = P(–1.75 < z < 1.75) = .9599 – .0401= .9198
(4) P(–1.5 z 1.5) = .9332 – .0668 = .8664

d.P(x > 136) = P(z > 2.25) = 1 – .9878 = .0122; 1.22%

6.29a.(1) P(x 959) = P(z 2.12) = .9830
(2) P(x > 1004) = P(z > 2.72) = 1 – .9967 = .0033
(3) P(x < 650) + P(x > 950) = P(z < –2) + P(z > 2) = .0228 + (1 – .9772) = .0456

1. P(x > 947) = P(z > 1.96) = .025

order = 947 boxes of cereal

6.30Restate each probability in terms of z. Then use the table to solve.

a.P(7 x 9) = P(–2.0 ≤ z ≤ 2.0) = .9772 – .0228 = .9544

b.P(8.5 x 9.5) = P(1 ≤ z ≤ 3) = .9987 – .8413 = .1574

c.P(6.5 x 7.5) = P(–3 ≤ z ≤ –1) = .1587 – .00135 = .15735

d.P(x 8) = P(z  0) = 1 – .5 = .5

e.P(x 7) = P(z  –2) = .0228

f.P(x 7) =P(z  –2) = 1 – .0228 = .9772

g.P(x 10) = P(z  4) = 1.0 (approximately)

h.P(x > 10) = P(z > 4) = 0 (approximately)

6.31a.P(x  27) = P(z  –3.00) = .5 – .00135 = .49865

b.Claim is probably not true, because the probability is very low of randomly purchasing a car with 27 mpg if the mean is actually 30 mpg.

6.32Common stocks: = 12.4,  = 20.6; let x = stock.
Tax free municipal bond: = 5.2,  = 8.6; let y = bond.

a.P(x > 0) = P(z > –.60) = 1 – .2743 = .7257

b.P(y > 0) = P(z > –.60) = 1 – .2743 = .7257

c.P(x > 10) = P(z > –.12) = 1 – .4522 = .5478

d.P(y > 10) = P(z > .56) = 1 – .7123 = .2877

e.P(x  –10) = P(z  –1.09) = .1379

f.P(y  –10) = P(z  –1.77) = .0384

6.33
= P(z < –2.14) + P(z > 1.94) = .0162 + (1 – .9738) = .0424

6.34P(xk) = .02


k = 31,784
Approximately 31,784 miles

Note: if use rounded z value of –2.05, then k =31,800

6.35a.10%, 90%

6.36
Solve the two equations for the two unknowns: = 75, = 12

6.37a.[± 2.33]

b.[50.575 ± 2.33(1.6438)] = [46.745, 54.405]

6.38a.+ 3= 5.2 + 3(8.6) = 31%

b.

6.39a.Process A:
Process B:
Process B is investigated more often.

b.Process A:
Process B:
Process A is investigated more often.

c.Process B will be investigated more often.

d.P(xk) = .3085 implies that . Thus k = 5000.
Investigate if cost variance exceeds $5000.

6.40P(xk) = .33


k = 2780
$2780

6.41
656 – = –.44896 – = 1.96
–= –.44 – 656–= 1.96– 896
= .44 + 656 = –1.96+ 896
.44 + 656 = –1.96 + 896
2.4 = 896 –656
2.4 = 240
= 100
= .44 + 646
= .44(100) + 656 = 44 + 656 = 700

6.42 Binomial tables are often unavailable for large values of n.

6.43Both np and n(1 – p) exceed 5.

6.44Create an interval that includes the integer (or integer interval) to be able to find the area under the curve. This is necessary because a discrete distribution is being approximated by a continuous distribution.

6.45a.np = (200)(.4) = 80

n(1 –p) = (200)(.6) = 120

both  5

b.Rounding z to 2 decimal places
(1) P(x = 80) = P(79.5 x 80.5) = P(–.072 x .072) = .0576.0558
(2) P(x 95) = P(x 95.5) = P(z 2.237) = .9874.9875
(3) P(x < 65) = P(x 64.5) = P(z –2.237) = .0126.0125
(4) P(x 100) = P(x 99.5) = P(z 2.8146) = .0024.0025
(5) P(x > 100) = P(x 100.5) = P(z 2.959) = .0015

6.46a.Both np and n(1 – p) exceed 5.

b.= 100,  = 7.0711
See the methods outlined in the solution to Exercise 5.45.
P(x = 80) = .001
P(x 95) = .2623
P(x < 65) = .0000
P(x 100) = .5282
P(x > 100) = .4718

6.47a.(1) Both np and n(1 – p) exceed 5.
(2)
(3) P(x 150) = P(x 150.5) = P(z –3.913) = less than .001

b.No. If the claim were true, the probability of observing this survey result is less than .001.

6.48a.
, rounding z to 2 decimal places gives .0104

b.No

6.49a.
(approximately)

b.No

6.50
P(xst) = .99


st = 441.7
442 units

6.51Explanations will vary.

6.52,

= 0 otherwise

6.53Explanations will vary.

6.54a. for x 0.

b.Graph not included in this manual.

c.

d.

e.

f.

g.

6.55a. for x 0.

b.Graph not included in this manual.

c.P(x 1) = .9502

d.P(.25 x 1) = .4226

e.P(x 2) = .0025

f.

g.

6.56a. for x 0.

b.Graph not included in this manual.

c.
(1) P(1 x2) = .2338
(2) P(x < 1) = .3729
(3) P(x > 3) = .2466
(4) P(.5 x 3.5) = .5966

d.

e.

6.57a. for x 0.

b.Graph not included in this manual.

c.
(1) P(x 3) = .8647
(2) P(1 x 2) = .2498
(3) P(x > 4) = .0695
(4) P(x < .5) = .2835

6.58a.

b. for x 0.

c.Graph not included in this manual.

d.

e.

f.Probably not; the probability of this result is quite small (.0198) if the claim is true.

6.59a.
(1)
(2)
(3)

b.Probably not; the probability of this happening is .2212 (which is not terribly small).

6.60See the steps on Pg 6-34?????? titled Normal Probability Plots

6.61That the data approximately follows a normal distribution

6.62a & b.

c.Data is skewed because plot does not show up as a straight line.

6.63Plot below shows data does not follow a normal distribution since it is not a straight line.

6.64Data is approximately normal because plot is close to a straight line.

6.65

6.66a.

b.P(x  74) = P(z  .5) = 1 – .6915 = .3085

c.P(x < 70.5) = P(z < –3) = .00135

6.67a.

b.

6.68a.

b. f(x)

1

–.5 .5 cents

c.P({x > .3} or {x < –.3}) = .2 + .2 = .4

1. P({x > .1} or {x < =.1}) = .4 + .4 = .8

f.

6.69a.P(x  3.5) = P(z  –1.25) = 1-.1056 = .8944

b.P(x  6) = P(z  0.83) = .7967

c.P(3.5  x  6) = .7967 – .1056 = .6911

6.70a.10%, 90%, approximately 3.462

Note: if use z = –1.28, k = 3.464

b.Follow the methods outlined in part a, or compute the inverse cdf in MINITAB.

Q1 = 4.196, Q3 = 5.804

6.71

Set lowest passing score to 298

6.72P(x < 15) = .004


= 15.053
Set to 15.053 inches.

6.73

6.74

a.

b.

6.75a, b. The probabilities below were computer calculated. Answers obtained using the normal table may be slightly different.

c. The probability of a return greater than 50% is:

(1)essentially zero for fixed annuities, cash equivalents, U.S., treasury bonds, U.S. investment grade corporate bonds, non-U.S. government bonds, and domestic large cap issues.

(2)greater than 1% for international equities, domestic MidCap stocks, and domestic small cap stocks.

(3)greater than 5% for domestic small cap stocks.

d. The probability of a loss is:

(1)essentially zero for fixed annuities and cash equivalents.

(2)

(3)greater than 1% for all investment-classes except fixed annuities and cash equivalents.

(4)greater than 10% for all investment classes except fixed annuities, cash equivalents, and U.S. treasury bonds.

(5)greater than 20% for domestic large cap stocks, international equities, domestic MidCap stocks, and domestic small cap stocks.

6.76

6.77

6.78x = 60 sec,  = 1

a.

b.

6.79 = 4.15,  = .5

a.

b.

c.

6.80 = np = (400)(.50) = 200

a.

b.Yes

6.81

6-1