Solutions to Practice Test 1

Each of the following problems is worth the same number of points. However, parts within a problem maynot be worth the same numbers of points. Put your final answers in the blanks provided if there are any.Show your work for partial credit.

1. Which of the following is not a property of a histogram?c. It uses a variable that is inherently categorical.(a, b, d, and e are true)

2. Which of the following is the least robust in a data set with outliers?c. Mode(The mode can rely on as few as two data values; all the others rely on more.)

3. Which of the following is not true of a discrete probability distribution?e. The values of y define events (subsets of the sample space) that may not be disjoint.

(The values define disjoint events.)

4. Which of the following is true for a uniform distribution?c. All of a, b, and c are true.(a, b and c are all properties of uniform distributions)

5. Which of the following is true for an outlier?e. All of a, b, and c can be true but don’t have to be.(Each of a, b, and c can be true but does not necessarily have to be true.)

6.. Given the following data set, find the information requested.

23, 34, 35, 46, 48, 52, 54, 56, 78, 100

a. the meanSolution:

y =

yii =1

n

∑n

=12 + 34 + L+ 100

10=

52610

= 52.6

b. the medianSolution:There are an even number of numbers (10) so it is the average of the two in the center, in this casethe fifth and the sixth number (when the numbers are in order from lowest to highest:

m =y n

2( ) +y n

2 +1( )2

=y 10

2( ) +y 10

2+1( )2

=y 5( )+ y 6( )

2=

48 + 522

= 50

c. the rangeSolution:The range is the largest value minus the smallest, so in this case:range = y(n ) − y(1) = 100 − 23 = 77

d. the standard deviationSolution:The standard deviation is the square root of the variance:

s =

yi − y ( )2

i =1

n

∑n − 1

=

yi2

i =1

n

∑ −

yii=1

n

∑

2

nn − 1

We can find that

yii =1

10

∑ = 526 while yi2

i =1

10

∑ = 32170 so

s =yi − y ( )2

i =1

n

∑n − 1

=yi

2

i =1

n

∑ −

yii=1

n

∑

2

nn − 1

=32170 −

526( )2

1010 − 1

=

=32170 − 276676

109

=32170 − 27667.6

9=

4502.49

= 500.266667 ≈

≈ 22.367

e. Find the interquartile rangeSolution:L =

14

n + 1( ) =14

10 + 1( ) =114

= 2.75 . We round up and QL = y 3( ) = 35 .

U =34

n + 1( ) =34

10 + 1( ) =334

= 8.25 . We round down and QU = y 8( ) = 56 .

So IQR = QU − QL = 56 − 35 = 21 .f. Find the z-score for 100.

Solution:z =

yi − y s

≈100 − 52.6

22.367=

47.422.367

≈ 2.12 .

g. Determine whether 100 is an outlier by the z-scoreSolution:Since z ≈ 2.12 = 2.12 < 3 , 100 is not an outlier (by the z-score method)

7. Given a random variable, y, which can take values 1, 2, 3,4, or 5, and withp(1)=.15,p(2)=.05, p(3)=.3, p(4)=.35, and p(5)=.15.

a. Find m.Solution:The problem above gives the following distribution:

y p y( )1 .152 .053 .34 .355 .15

We know that µ = E y( ) = y × p(y)( )all y∑ . One of the easiest ways to find this is to add a column to

the table:y p y( ) y × p y( )1 .15 .152 .05 .103 .3 .94 .35 1.45 .15 .75

Then we have µ = E y( ) = y × p(y)( )all y∑ = .15 + .10 + .9 + 1.4 + .75 = 3.3 .

b. Find s2.Solution:We know that σ 2 ≡ E y2( ) − µ 2 and we already know that µ = 3.3 so we only need to find E y2( ) .

We know that E y2( ) = y2 × p(y)( )all y∑ so, again, one of the easiest ways to find this is to add yet

another two columns to the table:y p y( ) y × p y( ) y2 y2 × p y( )1 .15 .15 1 .152 .05 .10 4 .203 .3 .9 9 2.74 .35 1.4 16 5.65 .15 .75 25 3.75

This gives E y2( ) = y2 × p(y)( )all y∑ = .15 + .20 + 2.7 + 5.6 + 3.75 = 12.4 .

So σ 2 ≡ E y2( ) − µ 2 = 12.4 − 3.3( )2 = 12.4 − 10.89 = 1.51.c. Find E(3y2+1).

Solution:One of the benefits of doing part (b) by finding E y2( ) is that it makes this part easy using theproperties of expected value:E 3y2 + 1( ) = E 3y2( ) + E 1( ) = 3E y2( ) + 1 = 3 12.4( ) + 1 = 38.2 .

8. Given a random variable y with a pdf

f (y) =

cy + 4 if 0 < y < 12

0 elswhere

a. Find the value of c that makes f (y) a legitimate pdf.Solution:We know that for a legitimate pdf that f (y)

−∞

∞

∫ dy = 1 . So we have

1 = f (y)−∞

∞

∫ dy = (cy + 4)0

12∫ dy

= cy2

2+ 4y

y=0

y= 12

=c 1

2( )2

2+ 4 1

2( )

−c 0( )2

2+ 4 0( )

= c8

+ 2

which means thatc8

+ 2 = 1

c8

= −1

c = −8

b. Find µy .Solution:We know that µy = E(y) and that E(y) = y f (y)

−∞

∞

∫ dy . So we have:

E(y) = y f (y)−∞

∞

∫ dy = y (−8y + 4)0

12∫ dy = (−8y2 + 4y)

0

12∫ dy

= −8 y3

3+ 4y2

2

y=0

y= 12

= − 8y3

3+ 2y2

y=0

y= 12

= −8 1

2( )3

3+ 2 1

2( )2

− −8 0( )3

3+ 2 0( )2

= −8 1

8( )3

+ 2 14( )

= − 13

+ 12

= − 26

+ 36

= 16

c. Find E y2( ) .Solution:We know that E(y2 ) = y f (y2 )

−∞

∞

∫ dy . So we have:

E(y2 ) = y2 f (y)−∞

∞

∫ dy = y2 (−8y + 4)0

12∫ dy = (−8y3 + 4y2 )

0

12∫ dy

= −8 y4

4+ 4y3

3

y=0

y= 12

= −2y4 + 4y3

3

y=0

y= 12

= −2 12( )4

+4 1

2( )3

3

− −2 0( )4 +4 0( )3

3

= − 216

+4 1

8( )3

= − 18

+ 16

= − 324

+ 424

= 124

d. Find σ y2

Solution:We know that σ y

2 = E(y2 ) − E(y)[ ]2 . So we have:

σ y2 = E(y2 ) − E(y)[ ]2

= 124

− 16

2

= 124

− 136

= 372

− 272

= 172

e. Find E 2y2 − 3y − 2( )Solution:

We know E(y) = 16

and E(y2 ) = 124

so

E 2y2 − 3y − 2( ) = E 2y2( ) + E −3y( ) + E −2( )= 2E y2( ) − 3E y( ) − 2

= 2 124

− 3 1

6

− 2 = 1

12− 1

2− 2

= 112

− 612

− 2412

= − 2912

≈ −2.41667

9. A student is taking a 10 question “true and false:” quiz. He doesn't know any of the answers so he takesout a coin and starts tossing it. When he gets a “heads,” he answers “true,” otherwise he answers“false.” The teacher is unfair and has made all of the answers on the quiz “true.” Unfortunately, thecoin the student chose is unbalanced and the probability of getting “Heads” is 0.3.

a. What is the probability that the student will get exactly seven questions right on the quiz?

Solution:This is a binomial situation with 10 trials so n = 10 and the probability of success ("Heads") isp = .3 . So the probability of failure is q = 1 − p = 1 − .3 = .7 . The probability of r successes in abinomial situation with n trials and probability of success p is:

We want to know the probability of r = 7 successes so we have:

p(7) =107

.3( )7 .7( )10 − 7 =

10!3!7!

.3( )7 .7( )3 =10 ⋅ 9 ⋅ 8

6.3( )7 .7( )3 ≈ 0.0090017

b. What is the expected number of questions he will get right?

Solution:For a binomial situation with n trials and probability of success p the mean is µ = E(y ) = npwhich gives us µ = E(y ) = np = 10( ) .3( ) = 3 so the expected number of questions he will get rightis 3.