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AP Statistics Solutions to Packet 7, Lecture notes of Statistics
Each die has six faces, marked with 1, 2, . . . 6 spots called pips. The dice used in casinos are carefully balanced so that each face is equally likely to ...
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HW #44 2, 3, 6 – 8, 13 – 17
7.2 THREE CHILDREN A couple plans to have three children. There are 8 possible arrangements of girls and boys. For example, GGB means the first two children are girls and the third child is a boy. All 8 arrangements are (approximately) equally likely.
(a) Write down all 8 arrangements of the sexes of three children. What is the probability of any one of these arrangements? BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG. Each has probability 1/8.
(b) Let X be the number of girls the couple has. What is the probability that X = 2? Three of the eight arrangements have two (and only two) girls, so P( X = 2) = 3/8 = 0.375.
(c) Starting from your work in (a), find the distribution of X. That is, what values can X take, and what are the probabilities for each value?
Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/
7.3 SOCIAL CLASS IN ENGLAND A study of social mobility in England looked at the social class reached by the sons of lower-class fathers. Social classes are numbered from 1 (low) to 5 (high). Take the random variable X to be the class of a randomly chosen son of a father in Class I. The study found that the distribution of X is:
Son’s class: Probability:
(a) What percent of the sons of lower-class fathers reach the highest class, Class 5? 1%
(b) Check that this distribution satisfies the requirements for a discrete probability distribution. All probabilities are between 0 and 1; the probabilities add to 1.
(c) What is P( X ≤ 3)? P( X ≤ 3) = 0.48 + 0.38 + 0.08 = 1 - 0.01 - 0.05 = 0.94.
(d) What is P( X < 3)? P( X < 3) = 0.48 + 0.38 = 0.86.
(e) Write the event “a son of a lower-class father reaches one of the two highest classes” in terms of X. What is the probability of this event? Write either X ≥ 4 or X > 3. The probability is 0.05 + 0.01 = 0.06.
(f) Briefly describe how you would use simulation to answer the question in (c). Read two random digits from Table B. Here is the correspondence: 01 to 48 ⇔Class 1, 49 to 86 ⇔Class 2, 87 to 94 ⇔Class 3, 95 to 99 ⇔ Class 4, and 00 ⇔Class 5. Repeatedly generate 2 digit random numbers. The proportion of numbers in the range 01 to 94 will be an estimate of the required probability.
7.8 VIOLENCE IN SCHOOLS, I An SRS of 400 American adults is asked. “What do you think is the most serious problem facing our schools?” Suppose that in fact 40% of all adults would answer “violence” if asked this question. The proportion ˆ p of the sample who answer “violence” will vary in
repeated sampling. In fact, we can assign probabilities to values of p ˆ using the normal density curve with mean 0.4 and standard deviation 0.024. Use the normal density curve to find the probabilities of the following events:
(a) At least 45% of the sample believes that violence is the schools’ most serious problem.
0.45 0.
P p ( ˆ 0.45) P Z P Z ( 2.083) 0.
− ≥
(b) Less than 35% of the sample believes that violence is the most serious problem.
P p ( ˆ < 0.35) = P Z ( < − 2.083) =0.
(c) The sample proportion is between 0.35 and 0.45.
P (0.35 ≤ p ˆ ≤ 0.45) = P ( 2.083 − ≤ Z ≤ 2.083) =0.
7.13 ROLLING TWO DICE Some games of chance rely on tossing two dice. Each die has six faces, marked with 1, 2,... 6 spots called pips. The dice used in casinos are carefully balanced so that each face is equally likely to come up. When two dice are tossed, each of the 36 possible pairs of faces is equally likely to come up. The outcome of interest to a gambler is the sum of the pips on the two up-faces. Call this random variable X.
(a) Write down all 36 possible pairs of faces. The 36 possible pairs of “up faces” are (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
(b) If all pairs have the same probability, what must be the probability of each pair? Each pair must have probability 1/36.
(c) Define the random variable X. Then write the value of X next to each pair of faces and use this information with the result of (b) to give the probability distribution of X. Draw a probability histogram to display the distribution.
Let X = sum of up faces. Then Sum Outcomes Probability x = 2 (1, 1) p = 1/ x = 3 (1, 2) (2, 1) p = 2/ x = 4 (1, 3) (2, 2) (3, 1) p = 3/ x = 5 (1, 4) (2, 3) (3, 2) (4, 1) p = 4/ x = 6 (1, 5) (2, 4) (3, 3) (4, 2) (5, 1) p = 5/ x = 7 (1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1) p = 6/ x = 8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2) p = 5/ x = 9 (3, 6) (4, 5) (5, 4) (6, 3) p = 4/ x = 10 (4, 6) (5, 5) (6, 4) p = 3/ x = 11 (5, 6) (6, 5) p = 2/ x = 12 (6, 6) p = 1/
7.16 HOW STUDENT FEES ARE USED Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an advisory board that represents student opinion. Suppose that 40% of all students oppose the use of student fees to fund student interest groups and that the opinions of the three students on the board are independent. Then the probability is 0.4 that each opposes the funding of interest groups.
(a) Call the three students A, B, and C. What is the probability that A and B support funding and C opposes it? (0.6)(0.6)(0.4) = 0.144.
(b) List all possible combinations of opinions that can be held by students A, B, and C. ( Hint: There are eight possibilities) Then give the probability of each of these outcomes. Note that they are not equally likely. The possible combinations are SSS, SSO, SOS, OSS, SOO, OSO, OOS, OOO (S = support, O = oppose). P(SSS) = 0.6^3 = 0.216, P(SSO) = P(SOS) = P(OSS) = (0.6^2 )(0.4) = 0.144, P(SOO) = P(OSO) = P(OOS) = (0.6)(0.4^2 ) = 0.096, and P(OOO) = 0.4^3 = 0.064.
(c) Let the random variable X be the number of student representatives who oppose the funding of interest groups. Give the probability distribution of X.
Value of X 0 1 2 3 Probability 0.216 0.432 0.288 0.
(d) Express the event “a majority of the advisory board opposes funding” in terms of X and find its probability. Write either X ≥ 2 or X > 1. The probability is 0.288 + 0.064 = 0.352.
7.17 A UNIFORM DISTRIBUTION Many random number generators allow users to specify the range of the random numbers to be produced. Suppose that you specify that the range is to be 0 ≤ Y ≤ 2. Then the density curve of the outcomes has constant height between 0 and 2, and height 0 elsewhere.
(a) What is the height of the density curve between 0 and 2? Draw a graph of the density curve. The height should be ½ since the area under the curve must be 1.
(b) Use your graph from (a) and the fact that probability is area under the curve to find P( Y ≤ 1)
P( Y ≤ 1) =
1 2 (c) Find P(0.5 < Y < 1.3). P(0.5 < Y < 1.3) = 0.
(d) Find P( Y ≥ 0.8). P( Y ≥ 0.8) = 0.
HW #45 ***2.28, 29, 31, 32 **7.22, 25, 26, 29
2.28 NORMAL REVIEW Use your calculator to find the proportion of observations from a standard normal distribution that falls in each of the following regions. In each case, sketch a standard normal curve and shade the area representing the region.
(a) z ≤ -2.25 P( z ≤ -2.25) = [normalcdf(-10, -2.25)] = 0.
(b) z ≥ -2.25 P( z ≥ -2.25) = [normalcdf( -2.25, 10)] = 0.
(c) z > 1.77 P( z > 1.77) = [normalcdf(1.77, 10)] = 0.
(d) -2.25 < z < 1.77 P(-2.25 < z < 1.77) = [normalcdf( -2.25, 1.77)] = 0.
2.29 MORE NORMAL REVIEW Use your calculator to find the value z of a standard normal variable that satisfies each of the following conditions. In each case, sketch a standard normal curve with your value of z marked on the axis.
(a) The point z with 70% of the observations falling below it. About 0.52 [invNorm(0.7)]
(b) The point z with 85% of the observations falling above it. About -1.04 [invNorm(0.15)]
(c) Find the number z such that the proportion of observations that are less than z is 0.8 About 0. [invNorm(0.8)]
(d) Find the number z such that 90% of all observations are greater than z. About -1. [invNorm(0.1)]
7.22 GRADE A DISTRIBUTION The table below gives the distribution of grades {A = 4, B = 3, and so on} in a large class:
Grade: Probability:
Find the average (that is, the mean) grade in this course. μ (^) X = (0)(0.10) + (1)(0.15) + (2)(0.30) + (3)(0.30)+ (4)(0.15) = 2.25.
7.25 KENO Keno is a favorite game in casinos, and similar games are popular with the states that operate lotteries. Balls numbered 1 to 80 are tumbled in a machine as the bets are placed, then 20 of the balls are chosen at random. Players select numbers by marking a card. The simplest of the many wagers available is “Mark 1 number.” Your payoff is $3 on a $1 bet if the number you select is one of those chosen. Because 20 of the 80 numbers are chosen, your probability of winning is 20/80, or 0.25.
(a) What is the probability distribution (the outcomes and their probabilities) of the payoff X on a single play? Value of X Probability:
The payoff is either $0 or $3.
(b) What is the mean payoff μ (^) X? For each $1 bet, μ (^) X = ($0)(.75) + ($3)(0.25) = 0.75.
(c) In the long run, how much does the casino keep from each dollar bet? The casino makes 25 cents for every dollar bet (in the long run).
7.26 GRADE DISTRIBUTION, II Find the standard deviation σ (^) X of the distribution of grades in
Exercise 7.22. In 7.22, we had μ = 2.25, so (^2) (0 2.25) (0.10) (^2) (1 2.25) (0.15) (^2) (2 2.25) (0.30) (^2) (3 2.25) (0.30) (^2) (4 2.25) (0.15) 2 σ (^) X = − + − + − + − + − = 1. σ (^) X = 1.3875 =1.
7.29 KIDS AND TOYS In an experiment on the behavior of young children, each subject is placed in an area with five toys. The response of interest is the number of toys that the child plays with. Past experiments with many subjects have shown that the probability distribution of the number X of toys played with is as follows: Number of toys xi : Probability pi :
(a) Calculate the mean μ (^) X and the standard deviation
μ (^) X = (0)(0.03) + (1)(0.16) + (2)(0.30) + (3)(0.23) + (4)(0.17) + (5)(0.11) = 2. 2 2 2 2 2 2 2
(5 2.68) (0.11) = 1.7176, and 1.7176 1.
X X
σ σ
(b) Describe the details of a simulation you could carry out to approximate the mean number of toys μ (^) X and the standard deviation σ (^) X. Then carry out your simulation. Are the mean and standard
deviation produced from your simulation close to the values you calculated in (a)? To simulate (say) 500 observations of x , using the T1-84, we will first simulate 500 random integers between 1 and 100 by using the command: randInt(1,100,500) →L The command sortA(L1) sorts these random observations in increasing order. We now identify 500 observations of x as follows:
Integers 1 to 3 correspond to x = 0 4 to 19 x = 1 20 to 49 x = 2 50 to 72 x = 3 73 to 89 x = 4 90 to 100 x = 5
For a sample run of the simulation, we obtained 12 observations of x = 0 86 x = 1 155 x = 2 118 x = 3 75 x = 4
- x = 5 These data yield a sample mean and standard deviation of (^) x = 2.64 and s =1. very close to μ and σ.
7.37 TIME AND MOTION, I A time and motion study measures the time required for an assembly-line worker to perform a repetitive task. The data show that the time required to bring a part from a bin to its position on an automobile chassis varies from car to car with mean 11 seconds and standard deviation 2 seconds. The time required to attach the part to the chassis varies with mean 20 seconds and standard deviation 4 seconds.
(a) What is the mean time required for the entire operation of positioning and attaching the part? The total mean is 11 + 20 = 31 seconds.
(b) If the variation in the worker’s performance is reduced by better training, the standard deviations will decrease. Will this decrease change the mean you found in (a) if the mean times for the two steps remain as before? No, changing the standard deviations does not affect the means.
(c) The study finds that the times required for the two steps are independent. A part that takes a long time to position, for example, does not take more or less time to attach than other parts. How would your answers to (a) and (b) change if the two variables were dependent with correlation 0.8? With correlation 0.3? The answers would not change. The total mean does not depend on dependence or independence of the two variables.
7.38 TIME AND MOTION, II Find the standard deviation of the time required for the two-step assembly operation studied in the preceding exercise, assuming that the study shows the two times to be independent. Redo the calculation assuming that the two times are dependent, with correlation 0.3. Can you explain in nontechnical language why positive correlation increases the variability of the total time. Assuming that the two times are independent, the total variance is (^2 2 2 22 42) 20, 4. σ (^) total = σ (^) pos + σ (^) att = + = so σtotal = 20 = seconds. Assuming that the two times are dependent with correlation 0.3, the total variance is (^2 2 2 2 2 42) 24.8,
total pos att pos att so (^) total seconds.
σ σ σ ρσ σ
σ
The positive correlation of 0.3 indicates that the two times have some tendency to either increase together or decrease together, which increases the variability of their sum.
7.41 Leona and Fred are friendly competitors in high school. Both are about to take the ACT college entrance examination. They agree that if one of them scores 5 or more points better than the other, the loser will buy the winner a pizza. Suppose that in fact Fred and Leona have equal ability, so that each score varies normally with mean 24 and standard deviation 2. (The variation is due to luck in guessing and the accident of the specific questions being familiar to the student.) The two scores are independent. What is the probability that the scores differ by 5 or more points in either direction?
If F and L are their respective scores, then F − L has a N ( 0, 22 + 22 )distribution, so
P(Fred scores 5 points more than Leona) + P(Leona scores 5 points more than Fred) =
5 0 0 5 ( 5) ( 5) 8 8 ( 1.7678) ( 1.7678) 2 ( 1.7678) 2(0.03854) 0.077 1
P F L P L F P Z P Z
P Z P Z P Z
HW #47 44, 49, 55 – 61, 62(skip a)
7.44 WEIRD DICE You have two balanced, six-sided dice. The first has 1, 3, 4, 5, 6, and 8 spots on its six faces. The second die 1, 2, 2, 3, 3, and 4 spots on its faces.
(a) What is the mean number of spots on the up-face when you roll each of these dice? First die: μ = (1)(1/ 6) + (3)(1/ 6) + (4)(1/ 6) + (5)(1/ 6) + (6)(1/ 6) + (8)(1/ 6) =4. Second die: μ = (1)(1/ 6) + (2)(1/ 6) + (2)(1/ 6) + (3)(1/ 6) + (3)(1/ 6) + (4)(1/ 6) = 2.
(b) Write the probability model for the outcomes when you roll both dice independently. From this, find the probability distribution of the sum of the spots on the up-faces of the two dice. 1 3 4 5 6 8 1 2 4 5 6 7 9 2 3 5 6 7 8 10 2 3 5 6 7 8 10 3 4 6 7 8 9 11 3 4 6 7 8 9 11 4 5 7 7 9 10 12
The probability distribution of X is:
x
P(X = x)
1 36
1 18
1 12
1 9
5 36
1 6
5 36
1 9
1 12
1 18
1 36
(c) Find the mean number of spots on the two up-faces in two ways: from the distribution you found in (b) and by applying the addition rule to your results in (a). You should of course get the same answer. (2)(1/ 36) (3)(1/18) (4)(1/12) (5)(1/ 9) (6)(5 / 36) (7)(1/ 6) (8)(5 / 36) (9)(1/ 9) (10)(1/12) (11)(1/18) (12)(1/ 36)
μ = + + + + +
- = 7 Using addition rule for means: μ = mean from 1st^ die + mean from 2nd^ die = 7.
7.49 A study of working couples measures the income X of the husband and the income Y of the wife in a large number of couples in which both the partners are employed. Suppose that you knew the
means μ (^) X and μ (^) Y and the variances σ (^) X^2 and σ (^) Y^2 of both variables in the population.
(a) Is it reasonable to take the mean of the total income X + Y to be μ (^) X + μY? Explain your answer.
Yes: This is always true; it does not depend on independence.
(b) Is it reasonable to take the variance of the total income to be σ (^) X^2 + σY^2? Explain your answer.
No: It is not reasonable to believe that X and Y are independent.
7.58 LIFE INSURANCE, IV The risk of insuring one person’s life is reduced if we insure many people. Use the result of the previous exercise and the rules for means and variances to answer the following questions.
(a) Suppose that we insure two 21-year-old males, and that their ages at death are independent. If X and Y are the insurer’s income from the two insurance policies, the insurer’s average income on the
two policies is: 0.5 0. 2
X Y
Z X Y
Find the mean and standard deviation of Z. You see that the mean income is the same as for a single policy but the standard deviation is less.
(^1 2 1 2 ) 4 4 2
Z 2 T X Z X Y X
μ = μ = μ = σ = σ + σ = σ =
(b) If four 21-year-old mean are insured, the insurer’s average income is
1 2 3 4
Z = X + X + X + X
Where Xi is the income from insuring one man. The Xi are independent and each has the same distribution as before. Find the mean and standard deviation of Z. Compare your results with the results of (a). We see that averaging over many insured individuals reduces risk. Using the new definition of Z , we have (^12) 4
Z Z Z 2 X
μ = unchanged and σ = σ = σ = smaller by a factor of
7.59 AUTO EMISSIONS The amount of nitrogen oxides (NOX) present in the exhaust of a
particular type of car varies from car to car according to the normal distribution with mean 1.4 grams
per mile (g/mi) and standard deviation 0.3 g/mi. Two cars of this type are tested. One has 1.1 g/mi of
NOX, the other 1.9. The test station attendant finds this much variation between two similar cars
surprising. If X and Y are independent NOX levels for cars of this type, find the probability
P X ( − Y ≥ 0.8 or X − Y ≤ −0.8)
that the difference is at least as large as the value the attendant observed.
X − Y is N(0, 0.3^2 + 0.3^2 ) = N(0, 0.4243),
so
P X Y P Z P Z
2[normalcdf(1.8856, 10)]
7.60 MAKING A PROFIT Rotter Partners is planning a major investment. The amount of profit X is uncertain but a probabilistic estimate gives the following distribution (in millions of dollars):
Profit: Probability:
(a) Find the mean profit μ (^) X and the standard deviation of the profit.
μ (^) X = (1)(0.1) + (1.5)(0.2) + (2)(0.4) + (4)(0.2) + (10)(0.1) = 3 million dollars. 2 σ X = (4)(0.1) + (2.25)(0.2) + (1)(0.4) + (1)(0.2) + (49)(0.1) = 6.35 million dollars So σ (^) X = 2.52 million dollars.
(b) Rotter Partners owes its source of capital a fee of $200,000 plus 10% of the profits X. So the firm actually retains Y = 0.9 X −0.
from the investment. Find the mean and standard deviation of Y. 0.9 0.2 2. 0.9 2.
Y X Y X
million dollars million dollars
μ μ σ σ
7.61 A BALANCED SCALE You have two scales for measuring weights in a chemistry lab. Both scales give answers that vary a bit in repeated weighings of the same item. If the true weight of a compound is 2.00 grams (g), the first scale produces readings X that have a mean of 2.000 g and standard deviation 0.002 g. The second scale’s readings Y have a mean of 2.001 g and standard deviation 0.001 g.
(a) What are the mean and standard deviation of the difference Y – X between the readings? (The readings X and Y are independent.)
2 2 2 2
Y X Y X Y X Y X Y X
g so g
μ μ μ σ σ σ σ
− 2 − −
(b) You measure once with each scale and average the readings. Your result is Z = (X + Y)/ 2. What are μ (^) Z and σ (^) Z? Is the average Z more or less variable than the reading Y of the less variable scale?
2 2 2
1 1 2 2 1 1 4 4
Z X Y
Z X Y Z
Y Z
g
so g
Z is slightly more variable than Y, so
μ μ μ
σ σ σ σ
σ σ