MATH534 Week 2 Homework

 

Answers to Review for Midterm #2

 

1)         True or false: The number of defective items, X, in a batch of 100 items is unknown prior to inspection. X is an example of a continuous random variable.

False.  Continuous means you can have decimal values, but “number of defectives” has to be a whole number (therefore discrete).

 

2)         True or false: In a distribution of a discrete random variable, X, the sum of the probabilities associated with the different possible values for X must always equal one.

This is true.  A probability distribution must always have the probabilities summing to 1.

 

3)         True or false: In a binomial experiment, each trial results in one of two possible outcomes which are dependent from trial to trial

This is false.  Yes, there are two outcomes (note the part of the word “bi”), but outcomes and trials are independent, not dependent.

 

4)         In a sample of ten students randomly selected from your class, the number of students who are right-handed is __________.

1. a) a sample random variable
2. b) a continuous random variable
3. c) a discrete random variable
4. d) a fixed variable

The answer is c —  Whole numbers make it discrete

 

5)         Consider the following discrete random distribution.

x                   P(x)

0                   .15

1                   .20

2                   .30

3                   .20

4                   .15

What is the mean of the distribution?

1. a) 5
2. b) 0
3. c) 5
4. d) 0

Take a look at the table on the next page.  The mean (expected value) of a distribution is the sum the sum of X * P(X) = 2.

 

 

 

X	P(X)	X*P(X)	(X-mean)^2*P(X)
0	0.15	0	0.6
1	0.20	0.2	0.2
2	0.30	0.6	0
3	0.20	0.6	0.2
4	0.15	0.6	0.6
	1.00	2	1.6
		Sum = mean	Sum = Variance

 

6)         What is the variance of the distribution above?

1. a) 0
2. b) 2
3. c) 5
4. d) 6

As seen in the table above, it is 1.6 (find how far each value is from the mean, then square, then multiply by the chance that it happens).  NOTE: the standard deviation is the square root of the variance.

 

7)         Thirty percent of the people called agree to participate in a brief telephone survey. In a random sample of four people called, what is the probability that exactly two people would not agree to participate in the survey?

1. a) 3000
2. b) 1500
3. c) 2646
4. d) 1698

P(2 don’t agree) = ₄C₂(.7)²(1-.7)²  = 6 * .49 * .09 = .2646

NOTE: ₄C₂ = 4!/[2!*(4-2)!] = (4*3*2*1)/(2*1)*(2*1) = 6

 

8)         What is the variance of a binomial distribution in which the number of trials is 100 and the probability of success is 0.4?

1. a) 240
2. b) 00
3. c) 00
4. d) 00

For a general probability distribution you’d have to find the variance like you did in example 6.  But for a binomial, life is easier….σ² = n * π * (1-π) = 100 * .4 * .6 = 24.  Again, remember that the standard deviation is the square root of the variance.

 

9)         60% of all graduate students in the music department at a large university are women. If a sample of 100 music graduate students is selected at random, how many women do you expect to see in that sample?

1. a) 50
2. b) 60
3. c) 70
4. d) 80

Again, for a general probability distribution you’d have to find the mean (expected value) like you did in example 5.  But for a binomial, life is easier….µ = n * π = 100 * .6 = 60.

 

10)       One fair coin is tossed 15 times, what is the probability of getting 3 heads out of 15 tossing experiments?

1. a) 01389
2. b) 01758
3. c) 125
4. d) 5

This is a binomial, so ₁₅C₃(.5)³(1-.5)¹²  = 455 * .125 * .000244 = .01389

NOTE: ₁₅C₃ = 15!/[3!*(15-3)!] = (15*14*13*12*11….)/(3*2*1)*(12*11*10*…) = 455

 

11)       According to the U.S. Census Bureau, 20% of the workers in Atlanta use public transportation. If 15 Atlanta workers are randomly selected…..

1. a) What is the expected number to use public transportation?

Expected value in a binomial is n*π = 15 * .2 = 3.  By the way, how do you know this is a binomial?  Because there are 2 outcomes for each of the workers in Atlanta (use public transportation or not), each worker is independent of the next, probability stays the same from trial to trial, etc.

 

1. b) What is the standard deviation for this distribution?

The standard deviation is the square root of n * π * (1-π) = sqrt(15*.2*.8) = sqrt(2.4) = 1.549

 

1. c) What is the chance of finding 4 out of the 15 who use public transportation?

No, the answer is NOT 4/15.  It is a binomial and we need to reflect the fact that 4 use it and the other 15 don’t.

P(4 “successes” out of 15) = ₁₅C₄(.2)⁴(1-.2)¹¹  =1365 * .0016 * .0859 = .1876

 

NOTE: ₁₅C₄ = 15!/4!*(15-4)! = [15*14*13*12*11*10 * etc.]/(4*3*2*1) * (11*10 *etc)

 

1. d) Use Excel to form the probability distribution for the number who use public transportation. Does Excel’s figure match your answer to part c?

Remember…a probability distribution is the table of all values of the random variable along with their probabilities.  In the first column put numbers from 0 through 15 (all the possible values), and next to it put the binomial formula:

= BINOM.DIST(A2,15,0.2,FALSE).

And yes, as you can see below the probability of 4 DOES = .1876.

X=use public transp	P(X)		X	P(X)		X	P(X)
0	0.035184		5	0.103182		10	0.000101
1	0.131941		6	0.042993		11	1.15E-05
2	0.230897		7	0.013819		12	9.54E-07
3	0.250139		8	0.003455		13	5.51E-08
4	0.187604		9	0.000672		14	1.97E-09
						15	3.28E-11

1. e) Graph the distribution in part d and comment.

 

 

Highlight the data from the probability distribution, insert a scatter diagram (I know that seems weird), and finally change the graph to a column chart.

 

This is a very skewed distribution (skewed right).  The high point of the curve is at 3…that makes sense since on average, we would expect to find 3 out of 15 (20%) using it.  But if we expect 3, it would be very unlikely to find a lot of people out of the 15 using it.  That is, the chance of finding 8 or more is really low!

 

1. f) What is the chance of finding at most 2 who use public transportation.

Well, P(at most 2) = P(2 or 1 or 0).  If you want to do it by hand that’s fine, but since we already have the probability distribution, let’s use it! P(at most 2) = .230897 + .131941 + .135184 = .398023.  So there is about a 40% chance of finding at most 2 who use it.

 

1. g) What is the chance of finding at least 7 who use public transportation? If you actually DID find 7 who use it, what would you conclude?

Again, P(at least 7) = P(7 or 8 or 9 or 10 or 11 or 12 or 13 or 14 or 15).  Use the table in Excel!  Probability  = .018.  So there is less than a 2% chance of finding that many using public transportation….in other words, very unlikely.  So if you DO find it, you would have to conclude that the 20% figure given in the beginning of the problem can’t be right.  It must be a lot higher.  In real life terms, that means that if they had a program trying to encourage people to use public transportation, it must be working!

 

NOTE: if you wanted you could have done this as a complement.  The probability of 0 through 6 = .982, so the probability of 7 or more must be 1 – .982.

 

12)       According to the American Medical Association, about 36% of all U.S. physicians under the age of 35 are women. Your company has just hired eight physicians under the age of 35 and none is a woman. If a group of women physicians under the age of 35 want to sue your company for discriminatory hiring practices, would they have a strong case based on these numbers? Use the binomial distribution to determine the probability of the company’s hiring result occurring randomly and comment on the potential justification for a lawsuit. 

 

This is a binomial distribution…the person is a woman or a man, and each person’s gender is independent of the next one.  So find the probability of 0 women in 8 trials:

 

₈C₀(.36)⁰(1-.36)⁸  = 1 * 1 * .0281 = .0281.      NOTE:  ₈C₀ = 8!/[0!*(8-0)!] = 1

 

It is rare (2.81% chance) to get 0 women out of a group of 8 if there was no discrimination.  So if that actually happened, there is something odd going on here.  It can be used as evidence in a lawsuit (this is real stuff!).  NOTE: “rare” means that the probability is less than .05…that’s what I keep bringing up in class as a piece of really important logic.

 

13)       True or false: In a uniform distribution, the height remains constant over the range of values between a and b.

True.  A uniform distribution looks like a rectangle with constant height from low to high value of X

 

14)       True or false: To standardize any value of a normally distributed random variable, we first subtract the mean of the distribution from the value and then divide it by the variance of the distribution.

False.  We divide by standard deviation, NOT variance.  Yes, I know this one is tricky…always read true/false questions carefully!

 

15)       For continuous distributions, probabilities of outcomes occurring between two points are determined by __________.

1. a) the area under the distribution curve outside the two points
2. b) the area under the distribution curve between the two points
3. c) the height along the vertical axis at the two points
4. d) the width of the horizontal axis between the two points

Remember: the word “probability” = “proportion” = “area”

 

16)       Scores on a certain exam are normally distributed with a mean of 78 and a variance of 25. What is the z-value for a score of 90?

2. a) 4
3. b) 8
4. c) 0
5. d) 2

(90 – 78)/sqrt(25) = 2.4

 

17)       Scores on a certain exam are normally distributed with a mean of 78 and a variance of 25. What is the probability of obtaining score equal to or greater than 90?

1. a) 9918
2. b) 0082
3. c) 5082
4. d) 9982

Look up Z = 2.4 and you will find .9918, so 1 – .9918 = .0082

 

18)       Scores on a certain exam are normally distributed with a mean of 78 and a variance of 25. What is the score obtained by the top 5% of the students?

91. a) 58
92. b) 50
93. c) 25
94. d) 18

 

The Z point for the top 5% can be found by looking up .95 (which is 1 – .05) in the body of the normal table (in among the 4 digit numbers).  It is exactly in between 1.64 and 1.65, so you can use either one or 1.645.    (X – 78)/sqrt(25) = 1.645  If you solve you get 86.225…just use the closest value which is 86.25.  OR if you use (X – 78)/sqrt(25) = 1.65  you get 86.25 exactly.

 

19)       Which one of the following is not a characteristic of a normal distribution?

1. a) Continuous
2. b) Symmetrical about its mean
3. c) Unimodal
4. d) Discrete

Remember, the normal distribution is a continuous distribution, not discrete!

 

20)       If the area to the left of z = 1.06 in the z-distribution is 0.8554, the area to the left of z = −1.06 is __________.

1. a) 1446
2. b) 3554
3. c) − 0.3554
4. d) 8554

1 – .8554 = .1446

 

21)       M/PF Research, Inc. lists the average monthly apartment rent in some of the most expensive apartment rental locations in the United States. According to their report, the average cost of renting an apartment in Minneapolis is $951. Suppose that the standard deviation of the cost of renting an apartment in Minneapolis is $96 and that apartment rents in Minneapolis are normally distributed. If a Minneapolis apartment is randomly selected, what is the probability that the price is:

1. a) $1,000 or more?

Since the data are normally distributed, use the 3 step method to solve: draw the picture (I’m not going to do that here), find the z point, and find the area on the table.

(1000-951)/96 = .5104 = .51 = Z, so area = 1 – .6950 = .3050

1. b) Between $900 and $1,100?

(900-951)/96 = -.53 = Z, and (1100-951)/96 = 1.55 = Z    So area = .9394 – .2981 = .6413

1. c) Between $825 and $925?

(825-951)/96 = -1.31 = Z, and (925-951)/96 = -.27 = Z  So area = .3936 – .0951 = .2985

1. d) Less than $700?

(700-951)/96 = -2.61 = Z, so area = .0045

 

22)       Suppose the average speeds of passenger trains traveling from Newark, New Jersey, to Philadelphia, Pennsylvania, are normally distributed, with a mean average speed of 88 miles per hour and a standard deviation of 6.4 miles per hour.

 

1. a) What is the probability that a train will average less than 70 miles per hour?

(70-88)/6.4 = -2.81 = Z, so area = .0025

 

1. b) What is the probability that a train will average more than 80 miles per hour?

(80-88)/6.4 = -1.25 = Z, so area = 1 – .1056 = .8944

 

1. c) What is the probability that a train will average between 90 and 100 miles per hour?

(90-88)/6.4 = .31, and (100-88)/6.4 = 1.88    so area = .9699 – .6217 = .3482

 

23)       The U.S. Bureau of Labor Statistics releases figures on the number of full-time wage and salary workers with flexible schedules. The numbers of full-time wage and salary workers in each age category are almost uniformly distributed by age, with ages ranging from 18 to 65 years. If a worker with a flexible schedule is randomly drawn from the U.S. workforce…
a)         What is the probability that he or she will be between 25 and 50 years of age?

The uniform distribution looks like a rectangle…equal probability for all ages!  So the probability = little (desired) area divided by large area = (50-25)/(65-18) = .5319

1. b) What is the mean value for this distribution?

Mean = (high + low)/2 = (a + b)/2 = (18+65)/2 = 41.5

1. c) What is the height of the distribution?

Height = 1/(b-a) = 1/(65-18) = .0213

1. d) What is the standard deviation for this distribution?

(65 – 18)/sqrt(12) = 13.568

 

24)       True or false: If a population is uniformly distributed, the means of the samples of size 30 drawn randomly from this population are also uniformly distributed.

False.  According to the Central Limit Theorem, the means have a normal distribution even if the original distribution of the data is uniform.

 

25)       True or false: The standard deviation of the sample means is called the standard error of the mean.

True (by definition).

 

26)       The standard deviation of the sample means is called __________.

1. a) sample standard deviation
2. b) standard error of the mean
3. c) mean standard deviation
4. d) standard deviation of the population

 

27)       Assume that about 75% of the graduate students can pass a test of physical fitness. A random sample of 200 graduate students was selected, what is the probability that less than 70% of those graduate students will pass the physical fitness?

1. a) 5516
2. b) 0516
3. c) 9484
4. d) 4484

 

σ =

So = -1.63 = Z    Area = .0516

 

28)       According to Nielsen Media Research, the average number of hours of TV viewing per household per week in the United States is 50.4 hours. Suppose the standard deviation is 11.8 hours and a random sample of 42 U.S. households is taken.

1. a) What type of data are we dealing with? Nominal, ordinal, interval, or ratio?

Data are ratio since for EACH household the hours of TV viewing is a number with a valid 0 point meaning “lack of”

1. b) What is the probability that the sample average is more than 52 hours?

(52 – 50.4)/11.8/sqrt(42) = 1.6/1.82077953 = .88 = Z

So area = 1 – .8106 = .1894

47. c) What is the probability that the sample average is less than 47.5 hours?

(47.5 – 50.4)/11.8/sqrt(42) = -2.9/1.82077953 = -1.59

So area = .0559

1. d) What is the probability that the sample average is less than 40 hours?

(40 – 50.4)/11.8/sqrt(42) = -10.4/1.82077953 = -5.71  (yes, it CAN be that large)

So area = virtually 0 since it is so far out in the tail that there is 0 area farther out.

 

29)       According to a study by Decision Analyst, 21% of the people who have credit cards are very close to the total limit on the card(s). Suppose a random sample of 600 credit card users is taken.

 

1. a) What type of data are we dealing with?  Nominal, ordinal, interval, or ratio?

The data are nominal (EACH individual person is either close to the limit or not).

 

1. b) What is the probability that more than 150 credit card users are very close to the total limit on their card(s)?

 

σ =

So  = 2.41 = Z    So area = 1 – .9920 = .0080

 

30)       A Travel Weekly International Air Transport Association survey asked business travelers about the purpose for their most recent business trip. Nineteen percent responded that it was for an internal company visit. Suppose 950 business travelers are randomly selected.

 

1. a) What type of data are we dealing with? Nominal, ordinal, interval, or ratio?

Again, the data are nominal since EACH traveler’s purpose is either for internal company visit or not.

 

1. b) What is the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit?

σ =

So = 4.71 = Z    So area = virtually 1 – 1 = 0 since it is almost off the chart

1. c) What is the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit?

= -3.14 = Z, and   = .79 = Z

 

so area =.7852 – .00084 = .78436

 

1. d) What is the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit?

 

p₁= 133/950 = .14   p₂= 171/950 = .18

 

= -3.93 = Z, and = -.79 = Z

 

so area = .2148 – .00004 = .21476

 

31 – 32)  These are up to you!