HW03-Sol - Chapter 3 Homework Solutions. Professor Su. PDF

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Chapter 3 Homework Solutions. Professor Su....

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STAT 3325

Probability & Applied Statistics

Solutions for Homework 03: Distribution of Random Variables

The following problems are taken from OpenIntro Statistics, the Third Edition: 3.2, 3.4, 3.6, 3.10, 3.13, 3.14, 3.16, 3.18, 3.30, 3.32, 3.33, 3.35, 3.36 1. Problem 3.2: (a) Pr(Z > −1.13) = 1 − 0.1292 = 0.8708 = 87.08%. (b) Pr(Z < 0.18) = 0.5714 = 57.14%. (c) Pr(Z > 8) ≈ 0.

(d) As shown in the graph below: Pr(|Z| < 0.5) = Pr(−0.5 < Z < 0.5) = Pr(Z < 0.5) − Pr(Z < −0.5) = 0.6915 − 0.3085 = 0.3830

2. Problem 3.4: (a) Let X denote the finishing times of Men, Ages 30 - 34 and Y denote the finishing times of emphWomen, Ages 25 - 29. Then, X ∼ N (µ = 4313, σ = 583) and Y ∼ N (µ = 5261, σ = 807). (b) The Z scores can be calculated as follows: ZLeo

x−µ σ

ZMary

x−µ σ

4948 − 4313 = 1.09 583 5513 − 5261 = = 0.31 807 =

Leo finished 1.09 standard deviations above the mean of his group’s finishing time and Mary finished 0.31 standard deviations above the mean of her group’s finishing time. 1

(c) Mary ranked better since she she has a lower Z score indicating that her finishing time is relatively shorter. (d) For Leo, we have Pr(Z > 1.09) = 1 − Pr(Z < 1.09) = 1 − 0.8621 = 0.1379 = 13.79% (e) For Mary, we have Pr(Z > 0.31) = 1 − Pr(Z < 0.31) = 1 − 0.6217 = 0.3783 = 37.83% (f) Answer to part (b) would not change as Z scores can be calculated for distributions that are not normal. However, we could not answer parts (c)-(d) since we cannot use the Z table to calculate probabilities and percentiles without a normal model. 3. Problem 3.6: (a) The fastest 5% are in the 5th percentile of the distribution. The Z score corresponding to the 5th percentile of the normal distribution is approximately −1.64. Then, Z = −1.65 =

x − 4313 ⇒ x = −1.65 × 583 + 4313 = 3351 seconds. 583

The fastest 5% of males in this age group finished in less than 56 minutes. (b) The slowest 10% are in the 90th percentile of the distribution. The Z score corresponding to the 90th percentile of the normal distribution is approximately 1.28. Then, Z = 1.28 =

y − 5261 ⇒ y = 1.28 × 807 + 5261 = 6294 seconds. 807

The slowest 10% of females in this age group took 1 hour, 45 minutes or longer to finish. 4. Problem 3.10: (a) Let X represent the heights of 10 years old. Then, X ∼ N (µ = 55, σ6). We have Pr(X < 48) = Pr{Z < (48 − 55)/6} = Pr(Z < −1.17) = 0.1210. (b) In order to calculate the probability that a randomly chosen 10 year old is between 60 and 65 inches, we need to calculate Pr(X < 65) and Pr(X < 60) and take the difference of the two probabilities:   60 − 55 65 − 55 Pr(60 < X < 65) = Pr 1.56) = 1 − 0.9406 = 0.0594. Pr(X > 50) = Pr Z < 3.2 Roughly 6% of passengers incur overweight baggage fees. 6. Problem 3.14: (a) The Z score corresponding to the top 2% (or the 98th percentile) is approximately 2.05. Z = 2.05 =

132 − 100 132 − 100 ⇒ σ= = 15.6 points. σ 2.05

(b) The Z score corresponding to the top 18.5% (or the 81:5th percentile) is approximately 0.90. Thus, Z = 0.90 = 7. Problem 3.16:

220 − 185 220 − 185 = 38.9 mg/dl. ⇒ σ= 0.90 σ

First find

Pr(X > 2100) = Pr{(2100 − 1500)/300} = Pr(Z > 2) = 1 − 0.9772 = 0.0228. Next, Pr(X > 1900) = Pr(Z > 1.33) = 1 − 0.9082 = 0.0918. Thus we have Pr(X > 2100|X > 1900) = = = 8. Problem 3.18: 3

Pr(X > 2100 & X > 1900 Pr(X > 1900) Pr(X > 2100) Pr(X > 1900) 0.0228 ≈ 0.25. 0.0918

(a) In order to determine if these data follow a nearly normal distribution we can check if they follow the 68 − 95 − 99.7% Rule. For this we calculate the intervals x ¯ ± k s for k = 1, 2, and 3 standard deviations away from the mean and determine what percent of the data fall within these intervals: k 1 2 3

Interval 61.52 ± 1 × 4.58 = (56.94, 66.10) 61.52 ± 2 × 4.58 = (52.36, 70.68) 61.52 ± 3 × 4.58 = (47.78, 75.26)

Frequency 17 17 17

Percentage 17/25 = 68% 24/25 = 96% 25/25 = 100%

68% of the data are within 1 standard deviation of the mean, 96% are within 2 and 100% are within 3 standard deviations of the mean. Therefore, we can say that the data approximately follow the 68-95-99.7% Rule. (b) The distribution of the scores is unimodal and symmetric. The superimposed normal curve seems to approximate the distribution pretty well. Therefore we can say that the distribution is nearly normal. The points on the normal probability plot also seem to follow a straight line. The points appear to jump in increments in the normal probability plot; it seems likely that this is due to rounding in the data. 9. Problem 3.30: Let X be the number of households who decide to respond. X has a binomial distribution with number of trials n = 15, 000 and p = 0.09. We are interested in the probability Pr(X ≥ 1.500), and we use the normal approximation to the binomial to calculate this probability. The mean and standard deviation of this distribution are

σ=

p

µ = np = 15, 000 × 0.09 = 1, 350 p np(1 − p) = 15, 000 × 0.09 × 0.91 = 35.

Then, the probability can be calculated as follows:   1, 500 − 1, 350 = Pr(Z > 4.29) ≈ 0. Pr(X ≥ 1, 500) = Pr Z ≥ 35 With such large numbers applying the continuity correction of 0.5 would barely make a difference. 10. Problem 3.32: (a) Pr(at least one afraid) = 1 − Pr(non afraid) = 1 − (1 − 0.07)1 0 = 1 − 0.6017

= 0.3983 (b) Pr(exactly 2 afraid) =

10 2

× 0.072 × 0.938 = 45 × 0.072 × 0.938 = 0.1234.

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(c) Pr(at most 1 afraid) = Pr(none afraid) + Pr(1 afraid)   10 = 0.4840 + × 0.071 × 0.939 1 = 0.4840 + 0.3643 = 0.8483. (d) If the group of 10 teenagers in a tent is a random sample, then there is about 84.83% chance that at most one of them will be afraid of spiders, and hence about 15.17% chance that 2 or more teenagers will be afraid of spiders in any particular tent. Because this probability is somewhat large ( usually we use 0.05 as a somewhat arbitrary cutoff rule), he probably should not randomly assign the teenagers to their tents. 11. Problem 3.33: (a) Pr(1st has green eyes, 2nd does not) = 0.125 × 0.875 = 0.109.

(b) Let X be the number of children with green eyes. Then, using a binomial distribution with n = 2 and p = 0.125;   2 × 0.1251 × 0.8751 = 0.219. Pr(X = 1) = 1

(c) Using a binomial distribution with n = 6 and p = 0.125   6 × 0.1252 × 0.8754 = 15 × 0.1252 × 0.8754 = 0.137. Pr(X = 2) = 2 (d) Using a binomial distribution with n = 6 and p = 0.125, Pr(X ≥ 1) = 1 − Pr(x = 0) = 1 − 0.449 = 0.551. (e) Let X = trial at which the first success (green eyes) occurs. Then, using a geometric distribution with p = 0.125 Pr(X = 4) = 0.8753 × 0.125 = 0.084. p (f) For √ Binomial(n = 6, p = 0.75), it has µ = np = 6 × 0.75 = 4 and σ = np(1 − p) = 6 × 0.75 × 0.25 = 1.06. Since 2 is (2 − 4)/1.06 = −1.89 standard deviations below the expected number of brown eyed children, strictly speaking this would not be considered unusual, since | − 1.89| < 2, however, it should be noted that the Z score for this value is pretty close to 2, making this observation somewhat unusual. 12. Problem 3.35: First we need to calculate the possible amounts one can win or lose when the game is played 3 times. Let X be the number of wins. Then X ∼ Binomial(n = 3, p = 18/38).

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Thus Pr(3 wins) = Pr(X = 3) = Pr(2 wins, 1 loss) = Pr(X = 2) = Pr(1 win, 2 losses) = Pr(X = 1) = Pr(3 losses) = Pr(X = 0) =

   3  0 3 20 18 3 38 38    2  1 3 20 18 2 38 38    1  2 3 20 18 1 38 38    0  3 3 20 18 0 38 38

= 0.1063 = 0.3543 = 0.3936 = 0.1458

Next, obtain the values of Y • If you win 3 times, the total winnings is Y = 1 + 1 + 1 = 3.

• If you win twice but lose once the total winnings is Y = 1 + 1 − 1 = 1.

• If you win once and lose twice the total winnings is Y = 1 − 1 − 1 = −1.

• Lastly, if you lose three times the total winnings is Y = (−1) + (−1) + (−1) = −3. These are the possible values Y can take, and the probabilities with which Y takes these values are calculated above. Then, the probability model for Y is: Y Pr(Y )

−3 0.1458

−1 0.3936

1 0.3543

3 0.1063

13. Problem 3.36: (a) Since she is randomly guessing, probability of getting each question right is p = 0.25. Let X be the number of trials it takes for the first success to occur. Then, using a geometric distribution with p = 0.25 : Pr(X = 3) = 0.753−1 × 0.25 ≈ 0.1406. (b) Let X be the number of questions she answers correctly. Then, using a binomial distribution with n = 5 and p = 0.25,   5 × 0.253 × 0.752 = 0.0879 Pr(X = 3) = 3   5 × 0.254 × 0.75 = 0.0146 Pr(X = 4) = 4 Pr(X = 3 or 4) = 0.0879 + 0.0146 = 0.1025

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(c) Majority means that she gets more than half of the questions right. Getting 3 or 4 or 5 of the questions right would all satisfy this condition. Then, Pr(majority right) = Pr(X ≥ 3) = Pr(X = 3) + Pr(X = 4) + Pr(X = 5)   5 = 0.0879 + 0.0146 + × 0.255 × 0.750 5 = 0.0879 + 0.0146 + 0.0010 = 0.1035.

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