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Introductory Statistics

David Diez, Mine Çetinkaya-Rundel, Christopher D Barr

Section A.4 Distributions Of Random Variables

Solution 1. Exercise 1
  1. 8.85\%.
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  2. 6.94\%.
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  3. 58.86\%.
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  4. 4.56\%. [Figure: A normal distribution centered at 0 where a smaller left tail of the distribution has been shaded at and below a location labeled -1.35.] [Figure: A normal distribution centered at 0 where a smaller right tail of the distribution has been shaded at and above a location labeled 1.48.] [Figure: A normal distribution centered at 0 where a central region has been shaded. The region that remains unshaded is a large left tail up to just below the mean and a small right tail also remains unshaded.] [Figure: A normal distribution centered at zero where the two tails below a value of -2 and above a value of 2 have been shaded.].
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Solution 2. Exercise 3
  1. Verbal: \(N(\mu = 151, \sigma = 7)\text{,}\) Quant: \(N(\mu = 153, \sigma = 7.67)\text{.}\)
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  2. \(Z_{VR} = 1.29\text{,}\) \(Z_{QR} = 0.52\text{.}\) [Figure: A normal distribution is shown along with 2 vertical lines specially marked. One is a little above the mean of the normal distribution at Z equals 0.52 and is labeled "QR". The second is a bit further above the mean at Z equals 1.29 and is labeled "VR"].
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  3. She scored 1.29 standard deviations above the mean on the Verbal Reasoning section and 0.52 standard deviations above the mean on the Quantitative Reasoning section.
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  4. She did better on the Verbal Reasoning section since her Z-score on that section was higher.
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  5. \(Perc_{VR} = 0.9007 \approx 90\%\text{,}\) \(Perc_{QR} = 0.6990 \approx 70\%\text{.}\)
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  6. \(100\% - 90\% = 10\%\) did better than her on VR, and \(100\% - 70\% = 30\%\) did better than her on QR.
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  7. We cannot compare the raw scores since they are on different scales. Comparing her percentile scores is more appropriate when comparing her performance to others.
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  8. Answer to part.
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  9. would not change as Z-scores can be calculated for distributions that are not normal. However, we could not answer parts.
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  10. -.
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  11. since we cannot use the normal probability table to calculate probabilities and percentiles without a normal model.
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Solution 3. Exercise 5
  1. \(Z = 0.84\text{,}\) which corresponds to approximately 159 on QR.
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  2. \(Z = -0.52\text{,}\) which corresponds to approximately 147 on VR.
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Solution 4. Exercise 7
  1. \(Z = 1.2\text{,}\) \(P(Z \gt 1.2) = 0.1151\text{.}\)
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  2. \(Z= -1.28 \to 70.6\degree\)F or colder.
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Solution 5. Exercise 9
  1. \(N(25, 2.78)\text{.}\)
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  2. \(Z = 1.08\text{,}\) \(P(Z \gt 1.08) = 0.1401\text{.}\)
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  3. The answers are very close because only the units were changed. (The only reason why they differ at all because 28\degree C is 82.4\degree F, not precisely 83\degree F.).
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  4. Since \(IQR = Q3 - Q1\text{,}\) we first need to find \(Q3\) and \(Q1\) and take the difference between the two. Remember that \(Q3\) is the \(75^{th}\) and \(Q1\) is the \(25^{th}\) percentile of a distribution. Q1 = 23.13, Q3 = 26.86, IQR = 26. 86 - 23.13 = 3.73.
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Solution 6. Exercise 11
  1. No. The cards are not independent. For example, if the first card is an ace of clubs, that implies the second card cannot be an ace of clubs. Additionally, there are many possible categories, which would need to be simplified.
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  2. No. There are six events under consideration. The Bernoulli distribution allows for only two events or categories. Note that rolling a die could be a Bernoulli trial if we simplify to two events, e.g. rolling a 6 and not rolling a 6, though specifying such details would be necessary.
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Solution 7. Exercise 13
  1. \(0.875^2\times 0.125 = 0.096\text{.}\)
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  2. \(\mu=8\text{,}\) \(\sigma=7.48\text{.}\)
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Solution 8. Exercise 15
If \({p}\) is the probability of a success, then the mean of a Bernoulli random variable \(X\) is given by \(\mu = E[X] = P(X = 0) \times 0 + P(X = 1) \times 1\) \(= (1 - p) \times 0 + p\times 1 = 0 + p = p\)
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Solution 9. Exercise 17
  1. Binomial conditions are met: (1) Independent trials: In a random sample, whether or not one 18-20 year old has consumed alcohol does not depend on whether or not another one has. (2) Fixed number of trials: \(n = 10\text{.}\) (3) Only two outcomes at each trial: Consumed or did not consume alcohol. (4) Probability of a success is the same for each trial: \(p = 0.697\text{.}\)
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  2. 0.203.
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  3. 0.203.
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  4. 0.167.
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  5. 0.997.
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Solution 10. Exercise 19
  1. \(\mu = 35\text{,}\) \(\sigma = 3.24\text{.}\)
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  2. \(Z = \frac{45 - 35}{3.24} = 3.09\text{.}\) 45 is more than 3 standard deviations away from the mean, we can assume that it is an unusual observation. Therefore yes, we would be surprised.
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  3. Using the normal approximation, 0.0010. With 0.5 correction, 0.0017.
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Solution 11. Exercise 21
  1. \(1-0.75^3 = 0.5781\text{.}\)
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  2. 0.1406.
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  3. 0.4219.
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  4. \(1-0.25^3=0.9844\text{.}\)
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Solution 12. Exercise 23
  1. Geometric distribution: 0.109.
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  2. Binomial: 0.219.
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  3. Binomial: 0.137.
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  4. \(1-0.875^6=0.551\text{.}\)
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  5. Geometric: 0.084.
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  6. Using a binomial distribution with \(n = 6\) and \(p=0.75\text{,}\) we see that \(\mu=4.5\text{,}\) \(\sigma=1.06\text{,}\) and \(Z = 2.36\text{.}\) Since this is not within 2 SD, it may be considered unusual.
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Solution 13. Exercise 25
  1. \(\stackrel{Anna}{1/5}\times\stackrel{Ben}{1/4}\times\stackrel{Carl}{1/3}\times\stackrel{Damian}{1/2}\times\stackrel{Eddy}{1/1} = 1/5!=1/120\text{.}\)
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  2. Since the probabilities must add to 1, there must be \(5!=120\) possible orderings.
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  3. \(8!=\text{40,320}\text{.}\)
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Solution 14. Exercise 27
  1. Geometric, 0.0804.
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  2. Binomial, 0.0322.
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  3. Negative binomial, 0.0193.
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Solution 15. Exercise 29
  1. Negative binomial with \(n=4\) and \(p=0.55\text{,}\) where a success is defined here as a female student. The negative binomial setting is appropriate since the last trial is fixed but the order of the first 3 trials is unknown.
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  2. 0.1838.
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  3. \({3 \choose 1} = 3\text{.}\)
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  4. In the binomial model there are no restrictions on the outcome of the last trial. In the negative binomial model the last trial is fixed. Therefore we are interested in the number of ways of orderings of the other \(k - 1\) successes in the first \(n - 1\) trials.
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Solution 16. Exercise 31
  1. Poisson with \(\lambda=75\text{.}\)
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  2. \(\mu=\lambda=75\text{,}\) \(\sigma=\sqrt{\lambda} = 8.66\text{.}\)
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  3. \(Z=-1.73\text{.}\) Since 60 is within 2 standard deviations of the mean, it would not generally be considered unusual. Note that we often use this rule of thumb even when the normal model does not apply.
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  4. Using Poisson with \(\lambda = 75\text{:}\) 0.0402.
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Solution 17. Exercise 33
  1. \(\frac{\lambda^k \times e^{-\lambda}}{k!} = \frac{6.5^5 \times e^{-6.5}}{5!} = 0.1454\text{.}\)
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  2. The probability will come to \(0.0015 + 0.0098 + 0.0318 = 0.0431\) (0.0430 if no rounding error).
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  3. The number of people per car is \(11.7 / 6.5 = 1.8\text{,}\) meaning people are coming in small clusters. That is, if one person arrives, there’s a chance that they brought one or more other people in their vehicle. This means individuals (the people) are not independent, even if the car arrivals are independent, and this breaks a core assumption for the Poisson distribution. That is, the number of people visiting between 2pm and 3pm would not follow a Poisson distribution.
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Solution 18. Exercise 35
0 wins (-\\(3): 0.1458. 1 win (-\\)1): 0.3936. 2 wins (+\\(1): 0.3543. 3 wins (+\\)3): 0.1063.
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Solution 19. Exercise 37
Want to find the probability that there will be 1,787 or more enrollees. Using the normal approximation, with \(\mu = np = 2,500 \times 0.7 = 1750\) and \(\sigma = \sqrt{np(1-p)} = \sqrt{2,500 \times 0.7 \times 0.3} \approx 23\text{,}\) \(Z = 1.61\text{,}\) and \(P(Z \gt 1.61) = 0.0537\text{.}\) With a 0.5 correction: 0.0559.
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Solution 20. Exercise 39
  1. \(Z=0.67\text{.}\)
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  2. \(\mu=\\)1650\(, \)x=\\(1800\text{.}\)
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  3. \(0.67 = \frac{1800-1650}{\sigma} \to \sigma=\\)223.88$.
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Solution 21. Exercise 41
  1. \((1-0.471)^2\times0.471 = 0.1318\text{.}\)
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  2. \(0.471^3 = 0.1045\text{.}\)
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  3. \(\mu = 1/0.471 = 2.12\text{,}\) \(\sigma=\sqrt{2.38} = 1.54\text{.}\)
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  4. \(\mu = 1/0.30 = 3.33\text{,}\) \(\sigma=2.79\text{.}\)
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  5. When \(p\) is smaller, the event is rarer, meaning the expected number of trials before a success and the standard deviation of the waiting time are higher.
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Solution 22. Exercise 43
\(Z = 1.56\text{,}\) \(P(Z \gt 1.56) = 0.0594\text{,}\) i.e. 6\%.
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Solution 23. Exercise 45
  1. \(Z = 0.73\text{,}\) \(P(Z \gt 0.73) = 0.2327\text{.}\)
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  2. If you are bidding on only one auction and set a low maximum bid price, someone will probably outbid you. If you set a high maximum bid price, you may win the auction but pay more than is necessary. If bidding on more than one auction, and you set your maximum bid price very low, you probably won’t win any of the auctions. However, if the maximum bid price is even modestly high, you are likely to win multiple auctions.
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  3. An answer roughly equal to the 10th percentile would be reasonable. Regrettably, no percentile cutoff point guarantees beyond any possible event that you win at least one auction. However, you may pick a higher percentile if you want to be more sure of winning an auction.
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  4. Answers will vary a little but should correspond to the answer in part.
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  5. . We use the 10\(^{th}\) percentile: \(Z = -1.28 \to \\)69.80$.
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Solution 24. Exercise 47
  1. \(Z = 3.5\text{,}\) upper tail is 0.0002. (More precise value: 0.000233, but we’ll use 0.0002 for the calculations here.).
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  2. \(0.0002 \times 2000 = 0.4\text{.}\) We would expect about 0.4 10 year olds who are 76 inches or taller to show up.
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  3. \({{2000}\choose{0}} (0.0002)^0 (1 - 0.0002)^{2000} = 0.67029\text{.}\)
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  4. \(\frac{0.4^0 \times e^{-0.4}}{0!} = \frac{1 \times e^{-0.4}}{1} = 0.67032\text{.}\)
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