Questions and Answers #2 Modeling Data Distributions

Questions and Answers Sheet 2 Modeling Data Distributions Question #1 Solve the following If the join probability distribution of X and Y iss given by 𝑥+𝑦 𝑓(𝑥, 𝑦) = 30 f or 𝑥 = 0, 2, 3: 𝑦 = 0, 1, 2 Find a) 𝑃(𝑋 ≤ 2, 𝑌 = 1); b) 𝑃(𝑋 > 2, 𝑌 ≤ 1); c) 𝑃(𝑋 > 𝑌); Answer: Step 1 Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved. Here, x 0 2 3 0 2 3 0 2 3 y 0 0 0 1 1 1 2 2 2 f ( x, y ) 0 0.066667 0.1 0.033333 0.1 0.133333 0.066667 0.133333 0.166667 Step 2 𝑎) 𝑃(𝑋 ≤ 2, 𝑌 = 1) = 𝑃(𝑋 = 0, 𝑌 = 1) + 𝑃(𝑋 = 2, 𝑌 = 1) = 𝑓(0,1) + 𝑓(2,1) = 0.033333 + 0.1 = 0.133333 Step 3 b) 𝑃(𝑋 > 2, 𝑌 ≤ 1) = 𝑃(𝑋 = 3, 𝑌 = 0) + 𝑃(𝑋 = 3, 𝑌 = 1) = 𝑓(3,0) + 𝑓(3,1) = 0.1 + 0.133333 = 0.233333 Step 4 c) 𝑃(𝑋 > 𝑌) = 𝑃(𝑋 − 𝑌 > 0) = 𝑃(𝑋 = 2, 𝑌 = 0) + 𝑃(𝑋 = 3, 𝑌 = 0) + 𝑃(𝑋 = 2, 𝑌 = 1) + 𝑃(𝑋 = 3, 𝑌 = 1) +𝑃(𝑋 = 3, 𝑌 = 2) = 0.066667 + 0.1 + 0.1 + 0.133333 + 0.166667 = 0.566667 Question #2 An electronics retailer offers an optional protection plan for a mobile phone it sells. Customers can choose to buy the protection plan for $100 and in case of an accident, the customer pays a $50 deductible and the retailer will cover the rest of the cost of that repair. The typical cost to the retailer is $200 per repair, and the plan covers a maximum of 3 repairs. Let X be the number of repairs a randomly chosen customer uses under the protection plan, and let F be the retailer's profit from one of these protection plans. Based on data from all of its customers, here are the probability distributions of X and F: Calculate the mean of X. X = # of repairs 0 1 2 3 F = retailer profit $100 −$50 −$200 −$350 Probability 0.90 0.07 0.02 0.01 Answer: Step 1 Given: X - number of repairs a randomly chosen customer uses under the protection plan F - retailer's profit from one of these protection plans. X 0 1 2 3 F 100 −50 −200 −350 Probability 0.9 0.07 0.02 0.01 Step 2 Mean of 𝑋 = summation of (𝑥 ⋅ 𝑝(𝑥)) = 0.14 X 0 1 2 3 F 100 −50 −200 −350 Probability 0.9 0.07 0.02 0.01 x  p( x) 0 0.07 0.04 0.03 Question #3 Prove the following Deviation boundary %of data points within boundary  2 3 68.0 95.40 99.70 Probability of any particular data point being outside boundary 32.0% 4.60% 0.30% Answer: Step 1 Empirical Rule : The empirical rule is used for forecasting final outcomes. After calculating the standard deviation and before collecting exact data, this rule can be used as a rough estimate of the outcome of the impending data to be collected and analyzed. Approximately 68% of the data lie within one standard deviation of the mean, that is, in the interval with endpoints 𝑥 ± 𝑠 for samples and with endpoints \mu\pm\sigma for populations; if a data set has an approximately bell-shaped relative frequency histogram, then approximately 95% of the data lie within two standard deviations of the mean, that is, in the interval with endpoints 𝑥 ± 2𝑠 for samples and with endpoints 𝜇 ± 2𝜎 for populations; and approximately 99.7% of the data lies within three standard deviations of the mean, that is, in the interval with endpoints 𝑥 ± 3𝑠 for samples and with endpoints 𝜇 ± 3𝜎 for populations. Step 2 One of the most necessary and sufficient condition for Empirical Rule are that the data distribution must be approximately bell-shaped and that the percentages are only approximately true. The Empirical Rule does not apply to data sets with severely asymmetric distributions, and the actual percentage of observations in any of the intervals specified by the rule could be either greater or less than those given in the rule. Question #4 Assume that a procedure yields a binomial distribution with a trial repeated 𝑛 = 5 times. Use some form of technology to find the cumulative probability distribution given the probability 𝑝 = 0.774 of success on a single trial. (Report answers accurate to 4 decimal places.) k P( X  k ) 0 1 2 3 4 5 Answer: Step 1 Use PH-stat to obtain the cumulative probability distribution for the given probability. Let X denotes the random variable which follows binimial distribution with the probability of succes 0.774 with number of trails selected is 5. That is, 𝑋 ∼ 𝐵(𝑛 = 5, 𝑝 = 0.774) The probability distribution is given by, n P( X = x) = ( ) p x (1 − p)n− x ; here 𝑥 = 0,1,2,⋅, 𝑛 for 0 ≤ 𝑝 ≤ 1 x Where n is the number of trials and p is the probability of success for each trial. Excel add-in (PHStat)procedure: 1) In EXCEL, Select Add-Ins > PHStat > Probability Prob. Distributions 2) Choose Binomial. 3) In Data enter Sample Size as 5 and Prob. of an Event on Interest as 0.774. 4) In Outcomes From and To, enter 0 and 5, respectively. 5) Select Cumulative Probabilities. 6) Click Ok. Excel add-in (PHStat) output: Data Sample size 5 Probability of an event of interest 0.774 Statistics Mean 3.87 Variance 0.8746 Standard deviation 0.9352 X 0 1 2 3 4 5 P ( X ) P ( X ) P ( X ) P (  X ) P ( X ) 0.0006 0.0006 0.0000 0.9994 1.0000 0.0101 0.0107 0.0006 0.9893 0.9994 0.0692 0.0798 0.0107 0.9202 0.9893 0.2368 0.3167 0.0798 0.6833 0.9202 0.4055 0.7222 0.3167 0.2778 0.6833 0.2778 1.0000 0.7222 0.0000 0.2778 Step 2 From the EXCEL output, the cumulative probabilities are as follows: K 0 1 2 3 4 5 P( X  k ) 0.0000 0.0006 0.0107 0.0798 0.3167 0.7222 Question #5 To find: Consider, X be the random variable that represents the number if adults who feel comfortable in a selfdriving vehicle. X 0 1 2 3 P( X ) 0.358 0.439 0.179 0.024 Answer: Step 1 A probability distribution function must satisfy the following conditions: 1) Sum of probabilities must be equal to 1. 2) Probabilities must lie between 0 and 1. Here, conditions required for a probability distribution are satisfied because probabilities are between 0 and 1 and sum of probabilities is equal to 1 which can be shown as: Sum of probabilities = 0.358 + 0.439 + 0.179 + 0.024 = 1 Step 2 Since, both the conditions for a probability distribution function are satisfied, the mean for the provided distribution can be calculated as: 𝐸(𝑋) = ∑ 𝑋 × 𝑃(𝑋) = 0(0.358) + 1(0.439) + 2(0.179) + 3(0.024) = 0.869 Step 3 The standard deviation can be calculated as: 2 Standard deviation (𝑋) = √𝐸(𝑋 2 ) − (𝐸(𝑋)) = √∑(𝑋 2 × 𝑃(𝑋)) − (𝐸(𝑋)) 2 = √((02 × 0.385) + (12 × 0.439) + ⋯ + (32 × 0.024)) − (0.869)2 = √1.371 − 0.755 =0.785 Thus, the mean and standard deviation are 0.869 and 0.785 respectively. Question #6 To find: Here we are given the probability distribution of Male Color Blindness as, x 0 1 2 3 4 5 P( x) 0.659 0.287 0.050 0.004 0.001 0+ Answer: Step 1 Note that by the property of probability distribution, 𝑃(0) + 𝑃(1) + 𝑃(2) + 𝑃(3) + 𝑃(4) + 𝑃(5) = 1 But 𝑃(0) + 𝑃(1) + 𝑃(2) + 𝑃(3) + 𝑃(4) = 1.001 > 1 which is a contradiction and 𝑃(5) = 1 − (𝑃(0) + 𝑃(1) + 𝑃(2) + 𝑃(3) + 𝑃(4)) = 1 − 1.001 = −0.001 < 0 which not possible. So, in this problem we are not given the probability distribution. Hence, we can’t calculate mean and standard deviation. Question #7 Five males with as X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the X-linked genetic disorder. X 0 1 2 3 4 5 P( X ) 0.031 0.156 0.313 0.313 0.156 0.031 Answer: Step 1 Conditions for a probability distribution: 𝑝(𝑥) > 0 ∑ 𝑝(𝑥) = 1 Step 2 It is given that x is the number of children who inherit the X-linked genetic disorder. Since all the probabilities of a random variable x are greater than Zero and sum of the probabilities are equal to one. Thus, it is a probability distribution. 𝑝(𝑥) > 0 ∑𝑥 𝑝(𝑥) = 0.031 + 0.156 + 0.313 + 0.313 + 0.156 + 0.031 = 1 Step 3 The mean is calculated below: The mean of a probability distribution is 2.5. 𝐸(𝑥) = ∑ 𝑥 × 𝑝(𝑥) 𝑥 = [0(0.031) + 1(0.156) + 2(0.313) + 3(0.313) + 4(0.156) + 5(0.031)] = 0 + 0.156 + 0.626 + 0.939 + 0.624 + 0.155 = 2.5 Step 4 The variance can be obtained as follows: 2 𝑉(𝑥) = 𝐸(𝑥 2 ) − (𝐸(𝑥)) Step 5 Now consider, 𝐸(𝑥 2 ) = ∑𝑥 𝑥 2 𝑝(𝑥) = 0 + 12 (0.156) + 22 (0.313) + 32 (0.313) + 42 (0.156) + 52 (0.031) = 0.156 + 1.252 + 2.817 + 2.496 + 0.775 = 7.496 Step 6 2 𝑉(𝑥) = 𝐸(𝑥 2 ) − (𝐸(𝑥)) = 7.496 − (2.5)2 = 7.496 − 6.25 = 1.246 Question #8 Probability of a sale person’s success in making a sale is 0.25. Salesperson made 4 attempts. Assuming that each attempt is independent and by using Binomial Distribution. a) Construct the probability distribution b) Find E(X) c) Var(x) List out various assumptions of probability distribution function and identify the areas where is it used Answer: Step 1 The probability of sales person's success in making a sale is given as, 𝑝 = 0.25 Apply the binomial distribuction to find the probability distribution. 𝑃(𝑋) = 𝑛 𝐶𝑟 𝑝𝑟 𝑞 𝑛−𝑟 a) The probability distribution is determined as, 𝑃(𝑋 = 0) = 4 𝐶0 (0.25)0 (0.75)4 = 0.316 𝑃(𝑋 = 1) = 4 𝐶1 (0.25)1 (0.75)3 = 0.422 𝑃(𝑋 = 2) = 4 𝐶2 (0.25)2 (0.75)2 = 0.211 𝑃(𝑋 = 3) = 4 𝐶3 (0.25)3 (0.75)1 = 0.047 𝑃(𝑋 = 4) = 4 𝐶4 (0.25)4 (0.75)0 = 0.004 Step 2 b) The mean of the probability distribution is determined as, 𝐸(𝑋) = 0 × (0.316) + 2 × (0.211) + 3 × (0.047) + 4 × (0.004) = 1.001 c) The variance of the probability distribution is determined as, 𝑉𝑎𝑟(𝑋) = 𝑛𝑝(1 − 𝑝) = 4 × (0.25) × (1 − 0.25) = 0.75 Step 3 The assumptions and areas where the probability distributions are used, 1) For calculating the confidence intervals for parameters. 2)For calculating the critical regions for hypothesis tests. 3) It is also used to determine a reasonable distributional model for the univariate data. 4) Statistical intervals and hypothesis tests are based on specific distributional assumptions. Before calculate an interval or test based on a distributional assumption, we need to verify that the assumption is justified for the given data set. In this type of intervals and tests, the distribution does not need to be the best-fit for the data, but sufficient enough so that the statistical intervals and hypothesis tests gives valid conclusions. Question #9 Let m1 and m2 denote the true average ages (in years) of students at two different colleges. Assume both population distributions are normal with equal variances. Calculate and interpret a 99% confidence interval for 𝑚1 − 𝑚2 using the following data. Sample size Sample mean Sample variance College 1 10 25 5 College 2 9 24 9 Group of answer choices a) 99% 𝐶. 𝐼. is (−2.10, −0.316) The samples do not suggest that the two colleges' students have significantly different ages. b) 99% 𝐶. 𝐼. is (−2.10, −0.316) The samples suggest that the two colleges` students have significantly different ages. c) 99% 𝐶. 𝐼. is (−2.10, −0.316) The samples suggest that the two colleges` students have significantly different ages. d) 99% 𝐶. 𝐼. is (−2.10, −0.316) The samples do not suggest that the two colleges` students have significantly different ages. Answer: Step 1 The 100(1 − 𝛼)% confidence interval for the difference between population means 𝜇1 − 𝜇2 , when the population standard deviations, 𝜎1 , 𝜎2 are unknown and the sample standard deviations, 𝑠1 , 𝑠2 are used, is the t-confidence interval as follows: (𝑥 ̅̅̅1 − ̅̅̅ 𝑥2 − 𝑡𝛼/2,𝑛1+𝑛2−22 𝑠 𝑝√ 𝑥1 1 1 , ̅̅̅ + 𝑛1 𝑛2 − ̅̅̅ 𝑥2 + 𝑡𝛼/2;𝑛1+𝑛2−2 𝑠 𝑝√ 1 1) + 𝑛1 𝑛2 Here, 𝑡𝛼/2;𝑛1+𝑛2−2 is the critical value of the t-distribution with degrees of freedom of 𝑛1 + 𝑛2 − 2 above which, 100(𝛼/2)% or 𝛼/2 proportion of the observations lie, and below which, 100(1 − 𝛼 + 𝛼/2)% = 100(1 − 𝛼/2)% or (1 − 𝛼/2) proportion of the observations lie. The sample size, sample mean and sample standard deviation for ages of students of college 1 are is, 𝑛1 = 10, ̅̅̅ 𝑥1 = 25 and 𝑠1 = √5 , respectively. The sample size, sample mean and sample standard deviation for ages of students of college 2 are is, 𝑛2 = 9, ̅̅̅ 𝑥2 = 25, and 𝑠2 = √9 , respectively. Denote 𝜇1 , 𝜇2 as the population ages of students of college 1 and 2, respectively. The pooled standard deviation is calculated below: (𝑛1 − 1)𝑠12 + (𝑛2 − 1)𝑠22 𝑠𝑝2 = 𝑛1 + 𝑛2 − 2 (10 − 1)(5) + (9 − 1)(9) = 10 + 9 − 2 117 = 17 ≈ 6.882352941 𝑠𝑝 = √6.882352941 ≈ 2.623423897 The student’s t distribution is used as the population variance is not known. The degrees of freedom is 17 = (10 + 95 − 2) The desired confidence limit is 99% . Hence, 𝛼 = 1 − 0.99 = 0.01 Using the Excel formula = 𝑇. 𝐼𝑁𝑉. 2𝑇(0.01,17), the critical value is 𝑡0.005, 17 = 2.898 Using the calculations above, and the given information, the 99% confidence interval is: (𝑥 ̅̅̅1 − ̅̅̅ 𝑥2 − 𝑡𝛼/2,𝑛1+𝑛2−22 𝑠 𝑝√ 𝑥1 1 1 , ̅̅̅ + 𝑛1 𝑛2 − ̅̅̅ 𝑥2 + 𝑡𝛼/2;𝑛1+𝑛2−2 𝑠 = ((25 − 24) ± (2.898)(2.623423897) (√ 𝑝√ 1 1) + 𝑛1 𝑛2 1 1 + )) 10 9 = {1 ± (2.898)(2.623423897)(0.459468291)) = (1 ± 3.49319152) = (−2.49, 4.49) The 99\% confidence interval is (−2.49, 4.49) The confidence interval contains the value 0. Hence, the samples do not suggest that the two colleges' students have significantly different ages. Correct option is D. Question #10 Algebraic Modeling subject x y −3 −2 −1 0 1 2 3 4 4 4 4 4 12 36 108 324 27 9 3 What is the exponential regression of the data? y=? Answer: Step 1 Given a dataset where an exponential function 𝑦 = 𝑎𝑏 𝑥 where a and b are constants. are to be obtained. Taking logarithm on both sides of the above equation, log 𝑦 = log 𝑎 + 𝑥 log 𝑏 ⇒ 𝜐 = 𝐴 + 𝐵𝑥 where 𝜐 = log 𝑦 𝐴 = log 𝑎 𝐵 = log 𝑏 Hence, the exponential function is reduced to a linear function between \upsilon and x and can be solved by least square method. By principle of least squares, the normal equations for estimating A and B is given by, ∑ 𝜐 = 𝑛𝐴 + 𝐵 ∑ 𝑥 ∑ 𝜐𝑥 = 𝐴 ∑ 𝑥 + 𝐵 ∑ 𝑥 2 Step 2 The following is table is formed to calculate the value of A and B y 4 27 4 9 4 3 4 12 36 108 324 Total x  = log y x 2  −3 −0.8293 9 2.4879 −2 −0.3522 4 0.7044 −1 0.1249 1 −0.1249 0 1 2 3 4 4 0.6021 1.0792 1.5563 2.0334 2.5105 6.7249 0 0 1 1.0792 4 3.1126 9 6.1002 16 10.0420 44 23.4014 Hence, putting the values obtained from the table to the normal equations we get, 1) 6.7249 = 8𝐴 + 4𝐵 2) 23.4014 = 4𝐴 + 44𝐵 On solving the simultaneous equations, by method of comparison, it is obtained that, from (1) 6.7249−𝑎𝐵 3) 𝐴 = 8 from (2) 23.4014−55𝐵 4) 𝐴 = 4 Comparing (3) and (4) 6.7249−4𝐵 23.4014−44𝐵 = 8 4 Hence solving the above equation, it is obtained that, 𝐵 = 0.4771 ⇒ 𝑏 = 𝐴𝑛𝑡𝑖𝑙𝑜𝑔 𝐵 = 2.9999 Then putting the value of B in (3), it is obtained that, 𝐴 = 0.6021 ⇒ 𝑎 = 𝐴𝑛𝑡𝑖𝑙𝑜𝑔 𝐴 = 4.0004 Therefore, the required exponential regression of the data is 𝑦 = 4.0005(2.9999) 𝑥