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Continuous Probability Distributions Test 5 Knowledge Application Thinking Communication /10 /10 /9 /6 Knowledge & Understanding Total /35 [ /10] Multiple Choice Identify the choice that best answers the question [1mark each] 1. Which of the following describes the normal distribution? a. bimodal b. skewed c. asymmetrical d. unimodal Answer: D 2. Which of the following data could not have a normal distribution? a. heights of trees c. masses of football players b. waiting time between two calls d. amount of rainfall Answer: B 3. Predictions based on a sample taken from a normally distributed population are most reliable when a. the sample size is small c. the standard deviation is small b. the sample size is large d. the standard deviation is large Answer: B 4. For a sample from normally distributed population, which of the following statements is false? a. The distribution of frequencies in the sample data tends to follow the same bell-shaped curve as the underlying distribution. b. The same sample mean, 𝑥̅ , and sample standard deviation, s, provide estimates of the population parameters, 𝜇 and 𝜎 . c. The larger the sample, the more reliably it will reflect the population. d. Continuity correction should be used when continuous data are being modeled. Answer: D 5. For each of the binomialdistributions listed below is the normal distribution not a reasonable approximation? a. n = 50, p = 0.4 b. n = 40, p = 0.12 c. n = 75, p = 0.11 d. n = 40, p = 0.8 Answer: B np = (40)(0.12) = 4.8 < 5 ∴ not a reasonable approximation 6. Assuming that a normal distribution is a reasonable approximation for a binomial distribution, what value is used to approximate the standard deviation? a. √𝑛𝑝𝑞 b. np c. npq d. √𝑛𝑝 Answer: A 7. Find the mean of a set of data with a standard deviation of 4, given that a value of 73 has a z-score of 1.5. a. 79 b. 61 c. 67.5 d. 67 Answer: D 𝑥− 𝜇 73 − 𝜇 z= ⟹ 1.5 = ⟹ 73 – 𝜇 = 6 ⟹ 73 – 6 = 𝜇 ⟹ 𝜇 = 67. 𝜎 4 8. The graph of the distribution of the numbers of Canadian families is a. positively skewed with a relatively low modal value and an extended tail b. positively skewed with a relatively high modal value and an extended tail c. negatively skewed with a relatively low modal value and an extended tail d. negatively skewed with a relatively high modal value and an extended tail Answer: A. 9. Illustrate P(X < 8.5) for a normal distribution with 𝜇 = 10 and 𝜎 = 1. [2] Answer: Application [ /10] 10. Suppose the commuting time from North Bay to Barrie varies uniformly from 135 min to 180 min. 1 1 𝜇 = 16 ⟹ k = = = 0.0625. 𝜇 16 Random variable X = “waiting time between buses” P (13 < X < 17) = P (X < 17) – P (X < 13). = (1 – e– 0.0625(17)) – (1 – e– 0.0625(13)) = e– 0.0625(13) – e– 0.0625(17)≈ 9.8%. a. Draw the graph of the probability density function and label it. [1] 1 (180 – 135) ∙ h = 1 ⟹ h = ≈ 0.022. 45 b. What is the probability that the drive will take less than 160 min? [1] 1 5 P(X < 160) = (160 – 135) ∙ h = 25 ∙ = 45 9 or P(X < 160) = 160 – 135 180 – 135 = 25 45 5 = . 9 11. At a busy radio station, the average time between request calls is 125 s. What is the probability that the next call will occur after a wait of 4 minutes? [2] For exponential distribution, Solution 𝑷(𝑿 < 𝑥 ) = 𝟏 − 𝒆−𝒌𝒙 . 1 1 𝜇 = 125 ⟹ k = = = 0.008. 𝜇 125 Random variable X = “time between request calls” P (X > 240) = 1 – P (X < 240). = 1 – (1 – e– 0.008(240)) = e– 0.008(240) ≈ 14.7%. 12. A really tough trivia quiz has 50 multiple-choice questions, each with four possible answers. Use a normal approximation to calculate the probability of randomly guessing the correct answers for at least half questions. Solution Random variable X = “number of the correct answers for 50 multiplechoice questions”. X is binomially distributed with p = 0.25 and n = 50. Here, np = (50) (0.25) = 12.5 > 5 and nq = (50) (0.75) = 37.5 > 5. Therefore, the normal approximation should be reasonable. µ = np = 12.5 and 𝜎 = √𝑛𝑝𝑞 = √(50)(0.25)(0.75) ≈ 3.06. Using the normal approximation with continuity correction, we find that the required probability is 24.5 – 12.5 P (X > 24.5) = P (Z > 3.06 Using graphing calculator: ) = P (Z < 3.92)≈ 0% P (X > 24.5) = normalcdf (24.5, 1E99, 12.5, 3.06) ≈ 4.4 × 10– 5. 13. A manufacturer of cereal finds that the masses of cereal in the company’s 200-g packages are normally distributed with a mean of 200 g and a standard deviation of 16.3 g. a) What proportion of these boxes contain between 183.7 and 216.3 grams of cereal? [4] Solution Random variable X = “mass of cereal in a box”. P (183.7 < X < 216.3) = normalcdf (183.7, 216.3, 200, 16.3) ≈ 0.6827. P (183.7 < X < 216.3) ≈ 68.27%. b) What is the probability that a box selected at random contains more than 216.3 g of cereal? Solution P (X > 216.3) = normalcdf (216.3, 1E99, 200, 16.3) ≈ 0.1587. P (X > 216.3) ≈ 15.87%. c) How many of shipment of 120 boxes would you expect to contain between 178 g and 225 g of cereal? Solution P (178 < X < 225) = normalcdf (178, 225, 200, 16.3) ≈ 0.8489. 0.8489 × 120 ≈ 102 boxes. d) Determine the centered range of masses that you would expect 90% of these boxes of cereal to contain. Solution Upper Bound = invNorm (0.95, 200, 16.3) ≈ 226.81 Lower Bound = invNorm (0.05, 200, 16.3) ≈ 173.18 ∴ P (173.18 < X < 226.81) = 0.90. Communication [ /6] 14. Explain the difference between discrete and continuous probability distributions. Provide an example of each. [2] Answer: Discrete probability distribution is the probability distribution of a discrete random variable, which has separated countable number of possible values. The probabilities of these values could be given by a formula or a table. Example Random variable X = “the number of times a red face card appears in 5 trials of selecting a card from a deck of cards”. The probability distribution of X is given by the formula: 5 3 𝑘 23 5 − 𝑘 P(X = k) = ( ) ( ) ( ) , k = 0, 1, 2, 3, 4, 5. 26 𝑘 26 Continuous probability distribution is the probability distribution of a continuous random variable, which can get any value from some number interval. This probability distribution is given by a probability density function – a function with non-negative values the area between which and the x-axes equals 1. Example Random variable X = “the weight of a bag of cereal selected from a number of bags in a supermarket with average weight 500 g and standard deviation of 2 g.” The probability distribution of X is the normal distribution with 𝜇= 500 and 𝜎 = 2. 15. What is a continuity correction? What is the importance, or use, of the continuity correction? [2] Answer: In probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. The continuity correction comes up most often when we are using the normal approximation to the binomial distribution. This makes calculations easier. 16. If a student came up to you and said that P(X = 10) = 0.01, where X is a random variable which has a normal distribution, how would you respond? [2] Answer: “Stupid guy! The probability that a continuous random variable assumes a particular value is zero: P(X = 10) = 0”. Thinking & Problem Solving 17. The marks in a physics course have a mean of 79 and a standard deviation of 12. The marks in a chemistry course have a mean of 70 and a standard deviation of 18. Andrew had a mark of 90% in both classes. Why might you consider one of these marks to be better than the other? [4] Solution Determine Andrew’s z-score in the physics course: 𝑥 − 𝑥̅ 90 − 79 z= = ≈ 0.92. 𝑠 12 Determine Andrew’s z-score in the chemistry course: 𝑥 − 𝑥̅ 90 − 70 z= = ≈ 1.11 𝑠 18 Since 1.11 > 0.92, we can consider Andrew’s chemistry mark to be better than his physics mark. OR Physics: P(X < 90) = normalcdf (– 1E99, 90, 79, 12) ≈ 0.8212. Thus, Andrew’s mark in the physics course is at the 82nd percentile Chemistry: P(X < 90) = normalcdf (– 1E99, 90, 70, 18) ≈ 0.8667. P (X < 70) = P (Z < 1.11) = 0.8665 Thus, Andrew’s mark in the chemistry course is at the 86th percentile Therefore, we can consider Andrew’s chemistry mark to be better than his physics mark. 18. A class of 150 students took a final examination in mathematics. The mean score was 72% and the standard deviation was 14%. Determine the percentile rank of a score of 89%, assuming that the marks were normally distributed. [2] Solution 𝜇 = 72, 𝜎 = 14 P (X < 89) = normalcdf (– E99, 89, 72, 14)≈ 0.8877 ∴ A mark of 89% is in the 88th percentile. 19. The annual returns from a particular mutual fund are believed to be normally distributed. The following table lists the annual returns for this fund over a 20-year period. Year Return (%) Year 1 6.2 11 2 11.3 12 3 16.1 13 4 16.9 14 5 9.8 15 6 18.3 16 7 11.2 17 8 –12.1 18 9 19.2 19 10 17.6 20 Return (%) 5.4 26.0 13.5 24.2 – 1.5 1.4 15.7 11.8 1.9 17.8 a) Determine the mean and standard deviation of the annual returns for this fund. [1] Answer: 𝑥̅ = 11.5, s = 9.2. b) What is the probability that an annual return will be (i) at least 9%? [1] Answer: P(X > 9) = normalcdf (9, 1E99, 11.5, 9.2) ≈ 0.6071 (ii) negative? Answer: P(X < 0) = normalcdf (–1E99, 0, 11.5, 9.2) ≈ 0.1056 c) For how many of the next ten years would you expect the fund to have an annual return greater than 6%? What assumptions are necessary to answer this question? [1] Solution P(X > 6) = normalcdf (6, 1E99, 11.5, 9.2) ≈ 0.7250 0.7250 × 10 = 7.25 yrs Assumption: the trend continues (stays same)