Questions and Answers #1 Data Distributions

Questions and Answers Sheet 1 Data Distributions Question #1 a) By Chebychev's Theorem, at least how many students in a class of 200 would score within the range 𝜇 ± 2𝜎? b) By Empirical Rule, how many tudents in a class of 200 would score within the range 𝜇 ± 2𝜎? Answer: a) According to chebychev's theorem, for any population with mean 𝜇 and standart deviation 𝜎: At least 75% of observations will lie within 𝜇 ± 2𝜎 At least 88.9% of observations will lie within 𝜇 ± 3𝜎 Suppose we have a class of 200 students. Then, according to Chebychev's theorem, 75% of 200 students will lie within a range of 𝜇 ± 2𝜎. That is, The number of students that lie within a range of 𝜇 ± 2𝜎 = 75% of 200 = 200 × 0.75 = 150 That is, 200 students are within range of 𝜇 ± 2𝜎 b) Empirical says that for data from a normal distribution, About 66.26% will lie within 𝜇 ± 1𝜎; About 95.44% will lie within 𝜇 ± 2𝜎: About 99.73 will lie within 𝜇 ± 3𝜎: Suppose we have a class of 200 students. Then, according to Empirical rule, 95.44% of 200 students will lie within a range of 𝜇 ± 2𝜎. That is, The number of students that lie within a range of 𝜇 ± 2𝜎 = 95.44% of 200 = 200 × 0.9544 = 190.88 ≈ 191 That is, 191 students are within range of 𝜇 ± 2𝜎 = Question #2 Suppose that the cumulative distribution function of the random variable x is x0 0,  F ( x) = 0.25 x, 0  x  5 1, 5 x  Determine the following: a) p(x<2.8) b) p(x>1.5) c) p(x < -z) d) p(x>b) Answer: Step 1 From the given information, The cumulative distribution function of the random variable x is. x0 0,  F ( x) = 0.25 x, 0  x  5 1, x5  Step 2 Determine P(X<2.8) That is, 𝑃(𝑋 < 2.8) = 𝐹(2.8) =0.25(2.8) =0.7 Step 3 Determine P(X>1.5) That is, 𝑃(𝑋 > 1.5) = 1 − 𝑃(𝑋 ≤ 1.5) =1-F(1.5) =1-[0.25(1.5)] =1-0.375 =0.625 Step 4 Determine P(X< -2) That is, P(X< -2)=F(-2) =0 Step 5 Determine P(X>6) That is, 𝑃(𝑋 > 6) = 1 − 𝑃(𝑋 ≤ 6) =1-F(6) =1-1 =0 Question #3 (True/False) The central limit theorem implies that: 1.a All variables have bell-shaped sample data distributions if a random sample contains at least 30 observations. 2.b. Population distributions are normal whenever the population size is large. 3.c. For large random samples, the sampling distribution of 𝑦 is approximately normal, regardless of the shape of the population distribution. 3.d. The sampling distribution looks more like the population distribution as the sample size increases. Answer: Answer: Hence options c and d is correct The central limit theorem implies that for large random samples, the sampling distribution is approximately normal, regardless of the shape of the population distribution. And the sampling distribution is looks like a population distribution as the sample size increases. Question #4 a. The range of WR scores that would contain about 95% of the scores in the drug dealer sample. Given info: The data have mount-shaped, symmetric distribution. n=100, 𝑥 = 39, 𝑠 = 6. b. The proportion of scores that lies above 51. c. The range of WR scores that contain nearly all samples of drug dealer sample. Answer: a) Calculation: Empirical rule: Empirical rule is a rule that can be applied to a data set with mound-shaped, symmetric frequency distributions. The rule says that approximately 68% measurements will lie between 1 standard deviation of the data, 95% will lie between 2 standard deviations and 99.7% will lie between 3 standard deviations of mean. The data have mound-shaped, symmetric distribution. Therefore, the empirical rule can be applied. Using empirical rule, 95% of observations lie between 2 standard deviations of mean. That is, the interval has 95% of observations that would be (𝑥 ± 2𝑠) Upper limit of interval: 𝑥 + 2 × 𝑠 = 39 + 2 × 6 = 51 Lower limit of interval: 𝑥 − 2 × 𝑠 = 39 − 2 × 6 = 27 Thus, the interval is (27,51). b) The distribution of data is symmetric. From part a, it is clear that 95% that is 0.95 observations lie between (27 and 51). Therefore, the remaining 0.05 observations will lie outside the interval. Since the distribution is symmetric, the proportion of observations, which lies below 27 and above 51, will be the same. 0.05 Thus, half of 0.05 observations will lie above. That is, = 0.025 observations will lie above 51. 2 c) Calculation: Empirical rule: Empirical rule is a rule that can be applied to a data set with mound-shaped, symmetric frequency distributions. The rule says that approximately 68% measurements will lie between 1 standard deviation of the data, 95% will lie between 2 standard deviations and 99.7% lie between 3 standard deviations of mean. The data have mount-shaped, symmetric distribution. Therefore, empirical rule can be applied. Using empirical rule, 99.7% of observations lie between 3 standard deviations of the mean that is almost all observations in the data. That is, interval has 97.9% of observations that would be (𝑥 ± 3𝑠). Upper limit of interval: 𝑥 + 3 × 𝑠 = 39 + 3 × 6 = 57 Lower limit of interval: 𝑥 − 3 × 𝑠 = 39 − 3 × 6 = 21 Thus, the interval is (21,57). Question #5 Fill in the missing value for the following percent frequency distribution. Annual Salaries Percent Frequency Under $30, 000 10 $30, 000 − 49,999 35 $50, 000 − 69,999 24 $70, 000 − 89,999 $90, 000 and over 5 Answer: Step 1 Percentage Frequency: A percentage frequency distribution is a display of data that specifies the percentage of observations that exist for each data point or grouping of data points. It is a particularly useful method of expressing the relative frequency of survey responses and other data. Many times, percentage frequency distributions are displayed as tables or as bar graphs or pie charts. The process of creating a percentage frequency distribution involves first identifying the total number of observations to be represented; then counting the total number of observations within each data point or grouping of data points; and then dividing the number of observations within each data point or grouping of data points by the total number of observations. The sum of all the percentages corresponding to each data is equal to 100. Step 2 Given data represents frequency distribution of annual salaries. It is stated that sum of percentage frequency is equal to 100. Let, x be the missing percent frequency. 10 + 35 + 24 + 𝑥 + 5 = 100 74 + 𝑥 = 100 𝑥 = 100 − 74 𝑥 = 26 Complete frequency table is shown below: Annual Salaries Percent Frequency Under $30, 000 10 $30, 000 − 49,999 35 $50, 000 − 69,999 24 $70, 000 − 89,999 26 $90, 000 and over 5 Question #6 Step 1 Days to Maturity for Short-Term Investments. Note that there are 40 observations, the smallest and largest of which are 36 and 99, respectively. Apply the preceding procedure to choose classes for limit grouping. Use approximately seven classes. Note: If in Step 2 you decide on 10 for the class width and in Step 3 you choose 30 for the lower limit of the first class, then you will get the same classes as used in Example; otherwise, you will get different classes (which is fine). TABLE Days to maturity for 40 short-term investments 70 62 75 57 51 64 38 56 53 36 99 67 71 47 63 55 70 51 50 66 64 60 99 55 85 89 69 68 81 79 87 78 95 80 83 64 39 86 98 70 Example Limit Grouping Days to Maturity for Short-Term Investments Table 2.6 displays the number of days to maturity for 40 short-term investments. The data are from BARRON’S magazine. Use limit grouping, with grouping by 10s, to organize these data into frequency and relative-frequency distributions. Answer: Answer may vary one of the possible solution is as follows: Limit value grouping: A grouping method in which each class is used to represents a possible range of values (class limits). Each range consists of smallest value as the lower limit of the class and largest value as the upper limit of the class. The class limits are termed as limit value classes, and this grouping method is referred to be limit value grouping. The most appropriate place for use this limit value grouping method is that the data to be whole number in nature with relatively more distinct observation to employ single value grouping. Steps to find the classes for the data set: Obtain the (approximated) number of class which is seven to fourteen is given in the information. Calculate the class width by using the following formula: class width = Maximum observation-Minimum observation Number of classes The minimum value is 36 and maximum value is 99. 99−36 class width = = 7 63 7 =9 Thus, the class width is approximately equal to 9. Obtain the lower class limit which is less than or equal to the minimum value of the data set. Thus the choice of lower limit of the class is 35 which is less than that of 36 (minimum observation). From the information given in table 2.6 of days to maturity for 40 short terms investment is started with class width of 10 then the first class limit will be less than 40 which is 35-44. Thus the remaining classes for the data set are given below: Days to maturity 35 − 44 45 − 54 55 − 64 65 − 74 75 − 84 85 − 94 95 − 104 99 Question #7 A privately owned liquor store operates both a drive-in facility and a walk-in facility. On a randomly selected day, let X and Y, respectively, be the proportions of the time that the drive-in and walk-in facilities are in use, and suppose that the joint density function of these variables is 2 𝑓(𝑥, 𝑦) = { (𝑥 + 2𝑦), 0 ≤ 𝑥 ≤ 1,0 ≤ 𝑦 ≤ 1, } 3 0, elsewhere. a. Find the marginal density of X. b. Find the marginal density of Y. c. Find the probability that the drive-in facility is busy less than one-half of the time. Answer: 1 𝑓(𝑥) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 0 1 2 𝑓(𝑥 + 2𝑦)𝑑𝑦 0 3 2𝑥 1 4 1 = ∫ 𝑑𝑦 + ∫ 𝑦𝑑𝑦 3 0 3 0 =∫ = = = 2𝑥 3 2𝑥 4 𝑦2 1 [𝑦]10 + [ ] 3 + 2 0 2 3 3 2(𝑥+1) 3 ,0 ≤ 𝑥 ≤ 1 b) The marginal density of Y is given below: 1 𝑓(𝑦) = ∫0 𝑓(𝑥, 𝑦)𝑑𝑥 12 = ∫0 (𝑥 + 2𝑦)𝑑𝑥 3 1 2 = ∫0 𝑥𝑑𝑥 + 3 2 𝑥2 1 = [ ] + 3 1 2 0 4𝑦 4𝑦 4𝑦 3 3 1 ∫0 𝑑𝑥 [𝑥]10 = + = 3 3 1+4𝑦 ,0 ≤ 𝑦 ≤ 1 3 c) It is given that drive in facility is denoted by x, then 1 1 𝑃 (𝑋 < ) = ∫02 𝑓(𝑥)𝑑𝑥 =∫ 2 1 2 2(𝑥 0 2 + 1) 𝑑𝑥 3 1 = ∫02(𝑥 + 1)𝑑𝑥 3 1 1 2 2 2 2 = ∫ 𝑥𝑑𝑥 + ∫ 𝑑𝑥 3 0 3 0 1 2 𝑥2 2 2 1 = [ ] + [𝑥]20 3 2 0 3 2 1 2 1 = [ ]+ [ ] 3 4 3 2 1 1 = + = 12 5 3 12 Therefore, the required probability is 0.4167. Question #8 Solve the following 1. If the joint probability distribution of X and Y is given by 𝑥+𝑦 𝑓(𝑥, 𝑦) = for x=0,2,3 ; y=0,1,2 30 Find a. 𝑃(𝑋 ≤ 2, 𝑌 = 1) b. 𝑃(𝑋 > 2, 𝑌 ≤ 1) c. P(X>Y) Answer: 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 a) 𝑃(𝑋 ≤ 2, 𝑌 = 1) = 𝑃(𝑋 = 0, 𝑌 = 1) + 𝑃(𝑋 = 2, 𝑌 = 1) = 𝑓(0,1) + 𝑓(2,1) =0.33333+0.1 =0.133333 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 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 #9 2 Is 𝐸(𝑋 2 ) equal to (𝐸(𝑋)) ? Answer: The expected value is the sum of the product of each possibility x with its probability (X=x), or as the formula suggests: 𝐸(𝑋) = 𝑠𝑢𝑚𝑥𝑃(𝑋 = 𝑥) Apply each of the suggested options to the formula 2 2 (𝐸(𝑋)) = (𝑠𝑢𝑚𝑥𝑃(𝑋 = 𝑥)) = (𝑠𝑢𝑚𝑥𝑃(𝑋 = 𝑥))(𝑠𝑢𝑚𝑥𝑃(𝑋 = 𝑥)) 𝐸(𝑋 2 ) = 𝑠𝑢𝑚𝑥 2 𝑃(𝑋 = 𝑥) 𝑠𝑢𝑚𝑥 2 𝑃(𝑋 = 𝑥)! = (𝑠𝑢𝑚𝑥𝑃(𝑋 = 𝑥))(𝑠𝑢𝑚𝑥𝑃(𝑋 = 𝑥)) Therefore, 𝐸(𝑋 2 )! = (𝐸(𝑋)) 2 Question #10 In an orchard, harvested apples are randomly distributed in packets, each containing 7 fruits. The orchard had a total of 25 apples collected, and 10 fruits were considered of excellent quality. Let X be the number of optimal apples allocated to first package, which of the probability distributions would be adequate to describe the random variable X? 7 (a) Geometric ( ) 25 3 (b) Binomial (7, ) 25 (c) Poisson (3) 3 (d) Geometric ( ) 25 (e) Hypergeometric (25,10,7) Answer: Step 1 Random Variable: In probability theory, a random variable is a variable whose values depend on outcomes of a random experiment. It is identified as a measurable function defined on a probability space that maps from a sample space to the real numbers. Discrete Random Variable: A discrete random variable is a variable whose values can be obtained by counting. Though the values can range infinitely, a discrete random variable is countably infinite. The number of students in a class and the number road accidents in a particular city are examples of discreate random variables. Probability Distribution: Let X be a random variable. Then, the probability distribution of X is defined as the statement of all possible values of X together with their corresponding probabilities. Step 2 Here, we have the total number of apples collected, i.e., 25, of which 10 fruits are of excellent quality. So, there are 25-10=15 fruits that are considered not of excellent quality. Also, apples are randomly distributed in packets, each containing 7 fruits, i.e., in the first box, 7 apples are distributed out of 25 apples. X is the number of optimal apples allocated to the first package. Obviously, this selection is without replacement as the same apples cannot be distributed more than once. So, there are 7-x apples allocated in the first package that are not of optimal quality. Therefore, the probability distribution of X will be given by,  10 15  ( x )(7 − x)  , x = 0,1, 2,...n 25 P( X = x) =  ( )  7  , otherwise 0 which implies that 𝑋 ∼ 𝐻𝑦𝑝𝑒𝑟 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐(25,10,7) Answer: The correct answer is: Option (e). Since the distribution of X is hypergeometric, it cannot be binomial, Poisson, or geometric. Hence, the other options are incorrect. Question #11 1. (4 marks) Check whether the mean and the variance of the following distribution exist: 𝑎 a. 𝑓𝑥 (𝑥) = 2 2) − ∞ < 𝑥 < ∞(𝑎 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) 𝜋(𝑎 +𝑥 b. 2 x −3 f x ( x) =  0 x 1 else 2. (5 mark) Let X be a uniformly distributed random variable with probability density function: a −1  x  3 f x ( x) =  0 else FInd the distribution for the random variable of 𝑌 = 𝑋 2 Answer: Step 1 Question 1: (a) Here 𝑓𝑥 (𝑥) is pdf of Cauchy distribution and we know that mean and variance does not exist for Cauchy distribution, since E|X| does not exist for this pdf. (b) Mean exist in this case, but variance does not exist as 𝐸|𝑋 2 | does not exist. Step 2 Question 2: The pdf of Y is  a  y  fY ( y ) =   a  2 y ,0  y 1 ,1  y  9 where a=1/4 Question #12 Random variables X and Y have the following joint probabilities: 𝑃(𝑋 = 0, 𝑌 = 0) = 𝑃(𝑋 = 0, 𝑌 = 8) = 1 𝑃(𝑋 = 16, 𝑌 = 0) = 𝑃(𝑋 = 16, 𝑌 = 16) = 4 a) Make a table of the joint distribution of X and Y and add their marginal distributions to the table. b) Compute Cov(X, Y). Are X and Y positively correlated, negative correlated, or uncorrelated? c) Compute the correlation coefficient between X and Y. Answer: Step 1 The random variables having joint probabilities are given as, 1 𝑃(𝑋 = 0, 𝑌 = 0) = 𝑃(𝑋 = 0, 𝑌 = 8) = 𝑃(𝑋 = 16, 𝑌 = 0) = 𝑃(𝑋 = 16, 𝑌 = 16) = 4 a. The joint distribution of X and Y is given as, X |Y 0 16 Fy (Y ) 0 1 4 1 4 1 2 8 16 Fx ( X ) 1 1 0 4 2 1 1 0 4 2 1 1 1 4 4 Step 2 b. The marginal distribution of X is given as, X P( x = X ) 0 16 1 1 2 2 𝐸(𝑋) = ∑𝑥 𝑥𝑃(𝑋 = 𝑥) 1 1 2 2 = 0 × + 16 × =8. 𝐸(𝑋 2 ) = ∑𝑥 𝑥 2 𝑃(𝑋 = 𝑥) 1 1 = 0 × + 162 × 2 2 =128. Step 3 The variance of the probability distribution is, 𝑉𝑎𝑟(𝑋) = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2 =128-64 =64 . The marginal distribution of Y is given as, Y P( y = Y ) 0 1 2 8 16 1 1 4 4 𝐸(𝑌) = 𝐸𝑥 𝑦𝑃(𝑌 = 𝑦) 1 1 1 = 0 × + 8 × + 16 × 2 4 4 =2+4 . =6 . Step 4 The value of 𝐸(𝑌 2 ) is, 𝐸(𝑌 2 ) = ∑𝑥 𝑦 2 𝑃(𝑌 = 𝑦) =0 1 1 1 = 02 × + 82 × + 162 × 2 4 4 =16+64. =80 . The variance of probability distribution is, 𝑉𝑎𝑟(𝑌) = 𝐸(𝑌 2 ) − [𝐸(𝑌)]2 =80-36 =44 Step 5 The covariance of the probability distribution is defined as, xy = 0 P( xy = XY ) 0 128 256 3 1 0 4 4 The value of E(XY) is, 3 1 𝐸(𝑋𝑌) = 0 × + 128 × 0 + 256 × 4 4 =64 . The covariance of X and Y is, Cov(X,Y), 𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸(𝑥𝑦) − 𝐸(𝑥) ⋅ 𝐸(𝑦) =64-48 =16 . Step 6 c. The correlation coefficient is defined as, 𝜌 = The value of V(x) is, 𝑉(𝑥) = 64. 𝑐𝑜𝑣(𝑥,𝑦) √𝑉(𝑥)⋅𝑉(𝑦) The value of Cov(x,y)=16. Step 7 Substitute the values, 16 𝜌= = = √64.44 16 √2816 16 53.066 = 0.3015 . Question #13 The joint density of X and Y is given by, 6 𝑥𝑦 𝑓𝑋𝑌 (𝑥, 𝑦) = (𝑥 2 + ) 0 < 𝑥 < 1,0 < 𝑦 < 2 7 2 a) Calculate 𝑃(𝑋 < 1, 𝑌 > 1) b) Find the marginal probability distributions of X and Y. c) Find the conditional probability density function of Y given X=0.5 and calculate 𝑃(𝑌 < 1/𝑋 = 0.5) Answer: Given: joint density of X and Y 6 𝑥𝑦 7 2 𝑓𝑥𝑦 (𝑥, 𝑦) = (𝑥 2 + ) 0 < 𝑥 < 1,0 < 𝑦 < 2 a) P(x<1, y>1) 1 𝑃(𝑥 < 1, 𝑦 > 1) = ∫ 𝑥=0 𝑃(𝑥 < 1, 𝑦 > 1) = P( x  1, y  1) = 1 ∫ [𝑥 2 𝑦 7 0 6 + 𝑥 𝑦2 2 6 2 𝑥𝑦 (𝑥 + ) 𝑑𝑦𝑑𝑥 2 𝑦=1 7 ∫ 2 ] 𝑑𝑥 2 2 1 6 1 2 3x 6 x3 3 x 2 1 6 17 102 ( x + ) dx = [ + ]0 =  = 7 0 4 7 3 4 2 7 24 168 =0.6071 𝑃(𝑥 < 1, 𝑦 > 1) = 0.6071 b) Marginal probability density of x and y. Marginal probability density of x 6 2 xy 6 x y2 2 6 ( x + )dy = [ x 2 y + ]0 = (2 x 2 + x) y =0 7 2 7 2 2 7 6 f x ( x) = (2 x 2 + x) 7 f x ( x) =  2 now, Marginal probability density of y 16 𝑓𝑦 (𝑦) = ∫0 (𝑥 2 + 7 2 3𝑦 7 14 𝑓𝑦 (𝑦) = + 𝑥𝑦 2 6 𝑥3 ) 𝑑𝑥 = [ 7 3 + 𝑦 𝑥2 1 2 3𝑦 7 14 ] = + 2 2 0 c) conditional density function of y given x=0.5 𝑦 𝑓(𝑥,𝑦) 𝑥 𝑓𝑥 (𝑥) 𝑓( ) = 𝑦 6 𝑓( )= 𝑥 = 0.5 7 xy (0.52 + y) xy 6 0.25 + 0.25 y 2 ( x 2 + ) (2 x 2 + x) = = 2 7 (2  0.5 + 0.5) 1.5 𝑓( 𝑦 𝑥=0.5 )= now, 𝑓 ( 0.25+0.25𝑦 𝑦<1 𝑥=0.5 1.5 1 ) = ∫0 6 2 𝑥𝑦 (𝑥 + ) 7 2 6 (2𝑥 2 +𝑥) 7 1 𝑑𝑦 = ∫0 0.25+0.25𝑦 1.5 [𝑦 + 𝑓( 𝑦2 𝑑𝑦 = 1 ] = 1.5 0.25 2 0 𝑦<1 𝑥=0.5 0.25 1.5 1 ∫0 (1 + 𝑦)𝑑𝑦 = 0.25 1.5 [1.5] = 0.25 ) = 0.25 Question #14 Martian potatoes begin to sprout very quickly after planting. Suppose X is the number of days after planting until a Martian potato sprouts. Then X has the following probability density function:  2 − x 3 −2x 1 −4x  e + e + e f ( x) =  7 14 14 0  ,0  x   otherwise a) What is the probability that X is within 1 standard deviation of its expected value? b) What is the probability that X = .6? Answer: Step 1 a) From the given information, the probability density function for X is,  2 − x 3 −2x 1 −4x  e + e + e f ( x) =  7 14 14 0  ,0  x   otherwise Step 2 ∞ Consider, 𝐸(𝑋) = ∫0 𝑥𝑓(𝑥)𝑑𝑥 ∞ 2 3 −𝑥 1 −𝑥 = ∫ 𝑥 ( 𝑒 −𝑥 + 𝑒2 + 𝑒 4 ) 𝑑𝑥 7 14 14 0 = −𝑥 ∞ 2 3 7 14 = ∫0 ( 𝑥𝑒 −𝑥 + 𝑥𝑒 2 + 1 14 −𝑥 𝑥𝑒 4 ) 𝑑𝑥 16 7 ∞ 𝐸(𝑋 2 ) = ∫ 𝑥 2 𝑓(𝑥)𝑑𝑥 ∞ 0 2 3 7 14 = ∫0 𝑥 2 ( 𝑒 −𝑥 + ∞ −𝑥 𝑒2 + 1 14 −𝑥 𝑒 4 ) 𝑑𝑥 2 3 2 −𝑥 1 2 −𝑥 = ∫ ( 𝑥 2 𝑒 −𝑥 + 𝑥 𝑒2 + 𝑥 𝑒 4 ) 𝑑𝑥 7 14 14 0 4 24 7 92 7 = + = + 64 7 7 Step 3 Therefore, 𝑆𝐷(𝑋) = √𝐸(𝑋 2 ) − [𝐸(𝑋)]2 =√ 92 7 16 2 −( ) 7 =2.8139 The probability that X is within 1 standard deviation of its expected value is 0.8810 and it is calculated below: 𝑃(𝐸(𝑋) − 𝑆𝐷 ≤ 𝑋 ≤ 𝐸(𝑋) + 𝑆𝐷) = 𝑃 ( = 𝑃(−0.52819 ≤ 𝑋 ≤ 5.099614) 16 7 − 2.8139 ≤ 𝑋 ≤ 16 7 − 2.8139) ≈ 𝑃(0 ≤ 𝑋 ≤ 5) 5 2 3 7 14 = ∫0 ( 𝑒 −𝑥 + 15 = 25 −𝑥 𝑒2 + 1 14 −𝑥 𝑒 4 ) 𝑑𝑥 5 −3𝑒 5 +7𝑒 2 −2𝑒 4 −2𝑒 2 15 7𝑒 2 =0.8810 b) Note: For continuous distributions, the probability at exact values is zero. Here, given density function is continuous distribution. Hence, the probability that X = 0.6 is 0. Question #15 Suppose that the random variable X is defined as 𝑋 = 0.6𝑋1 + 0.3𝑋2 + 0.1𝑋3 , where the probability distributions of the independent random variables 𝑋1 , 𝑋2 , and 𝑋3 are 5 f X 1 ( x) = ( )0.1x  0.95− x x 5 f X 2 ( x) = ( )0.4 x  0.65− x x 5 f X 3 (3) = ( )0.6 x  0.45− x x x=0,1,2,3,4,5 a) Evaluate the mean of X. b) Evaluate the variance of X. Answer: Step 1 Given: 𝑋 = 0.6𝑋1 + 0.3𝑋2 + 0.1𝑋3 Where all three are independent 5 5 f X 1 ( x) = ( )0.1x  0.95− x ][ f X 2 ( x) = ( )0.4 x  0.65− x ] x x 5 [ f X 3 (3) = ( )0.6 x  0.45− x x x=0,1,2,3,4,5 Step 2 It can be seen that all the three distributions follows binomial distribution So, Mean of binomial is 𝑛 ×𝑝 Variance of binomial is 𝑛 ×𝑝 ×𝑞 So, 𝑋1 ∼ 𝐵𝑖𝑛(5,0.1) 𝐸[𝑋1 ] = 5 × 0.1 = 0.5 𝑉𝑎𝑟[𝑋1 ] = 5 × 0.1 × 0.9 = 0.45 𝑋2 ∼ 𝐵𝑖𝑛(5,0.4) 𝐸[𝑋2 ] = 5 × 0.4 = 2 𝑉𝑎𝑟[𝑋2 ] = 5 × 0.4 × 0.6 = 1.2 𝑋3 ∼ 𝐵𝑖𝑛(5,0.6) 𝐸[𝑋3 ] = 5 × 0.6 = 3 𝑉𝑎𝑟[𝑋3 ] = 5 × 0.6 × 0.4 = 1.2 Step 3 a) Mean of X 𝐸[𝑋] = 0.6 × 𝐸[𝑋1 ] + 0.3 × 𝐸[𝑋2 ] + 0.1 × 𝐸[𝑋3 ] 𝐸[𝑋] = 0.6 × 0.5 + 0.3 × 2 + 0.1 × 3 E[X]=1.2 Hence, the mean is 1.2 Step 4 b) If the events are independent, then the variance is 𝑉𝑎𝑟[𝐴 + 𝐵] = 𝑉𝑎𝑟[𝐴] + 𝑉𝑎𝑟[𝐵] Here, all are independent, so the variance of X can be calculated as follows 𝑉𝑎𝑟[𝑋] = 0.62 × 𝑉𝑎𝑟[𝑋1 ] + 0.32 × 𝑉𝑎𝑟[𝑋2 ] + 0.12 × 𝑉𝑎𝑟[𝑋3 ] 𝑉𝑎𝑟[𝑋] = 0.62 × 0.45 + 0.32 × 1.2 + 0.12 × 1.2 𝑉𝑎𝑟[𝑋] = 0.282 The variance is 0.282. Question #16 You are given: (i) The annual number of claims on a given policy has the geometric distribution. (ii) One-third of the policies have two claims (𝛽 = 2), abd the remaining two-thirds have five claims (𝛽 = 5), Answer: Step 1 As per given: The annual number of claims on a given policy has the geometric distribution. And one third of the policies have two claims, and remaining two thirds have five claims. For the geometric distributions: 𝛽2 𝑃(𝑋1 = 2𝛽) = (1+𝛽)3 The expected value is 𝛽. Step 2 Solving it: 𝑃(𝛽 = 2|𝑋1 = 2) = 𝑃(𝛽 = 2|𝑋1 = 2) = 𝑃(𝑋1 𝑃(𝑋1 = 2|𝛽 = 2)𝑃(𝛽=2) = 2|𝛽 = 2)𝑃(𝛽=2)+𝑃(𝑋1 = 2|𝛽 = 5)𝑃(𝛽=5) 4 1 × 27 3 4 1 25 2 × + × 27 3 216 3 𝑃(𝛽 = 2|𝑋1 = 2) = 0.39024 Hence the expected value is: 𝛽 = (0.39024)(2) + (0.60976)(5) 𝛽 = 3.83 Question #17 The joint distribution of X and Y is as follows. Values of X Values of Y 1 1 0.1 0 0.3 P{Y = y} 0.4 a. Find the marginal distribution of X and Y. 0 P{ X = x} 0.2 0.3 0.4 0.7 0.6 1.0 b. Find the conditional distribution of Y given 𝑋 − 1, 𝑃{𝑌 = 𝑦|𝑋 = 1} c. Compute the conditional expectation of Y given 𝑋 = 1, 𝐸 {𝑌 = 𝑦|𝑋 = 1} Answer: a) Marginal distribution of X x 1 0 P( X = x) 0.3 0.7 Marginal distribution of Y x 1 0 P (Y = y ) 0.4 0.6 b) P(Y|X=1) 𝑃(𝑌 = 0|𝑋 = 1) = = 𝑃(𝑋 = 1, 𝑌 = 0) 𝑃(𝑋 = 1) 0.2 0.3 =0.6667 𝑃(𝑌 = 1|𝑋 = 1) = = 𝑃(𝑋=1,𝑌=1) 𝑃(𝑋=1) 0.1 0.3 =0.3333 c) E[Y|X=1] = ∑𝑦 𝑦 × 𝑃(𝑌 = 𝑦|𝑋 = 1) =0× 0.2 0.3 +1× 0.1 0.3 =0.3333 The required expectation is 0.3333. Question #18 The observations given below are related with duration between two patients that come to ER service of certain hospital. There is a claim that the distribution of durations is Exponential with V(X) = 1.778. Evaluate the claim. 3.264 4.62 0.439 0.13 1.975 1.271 0.82 0.42 1.145 0.61 2.408 0.87 1.942 0.32 0.62 0.375 0.383 2.512 1.92 3.19 2.12 3.89 Answer: Step 1 Under null hypothesis 𝐻0 : The distributions between two patients follow exponential distribution with mean 1.778 1 𝜇 = = 1.778 𝜆 Under alternate hypothesis 𝐻1 : The distributions between two patients do not follow exponential distribution of which mean is 1.778 1 𝜇 = ≠ 1.778 𝜆 Step 2 Sample Mean: 𝑥= ∑ 𝑥𝑖 𝑛 = 35.244 22 =1.602 1 Exponential distribution has mean and standard deviation 𝜇 = and 𝜎𝑥 = 𝜆 1 𝜆2 Using central limit theorem, 1 1 𝑋 ∼ 𝑁 ( , 2) 𝜆 𝜆 Step 3 Test statistic: 𝑧= = 𝑋−1/𝜆 1 √ 2 𝜆 1.602−1.778 √ = 1 1/1.7782 1.602−1.778 √1.7782 -0.09899 Computation of P-value:The P-value for the two tailed z-value test is 0.9211 which can be obtained using the excel formula, ''=2*NORM.S.DIST(-0.09899,TRUE)''. Step 4 Decision rule: If p-value ≤ 𝛼, then reject the null hypothesis. Otherwise, do not reject the null hypothesis. Conclusion: Consider level of significance as 0.05. The p-value is greater than level of significance. Hence, we have insufficient evidence to reject null hypothesis and thus conclude that the duration between two patients that come to ER service of certain hospital follow exponential distribution with mean 1.778. Question #19 Given the r.v. X with 0, elsewhere, pdff x ( x) =  cx, 0  x  1 Find (1) the value of c; (2) P(X<0.5) ; (3) the cdf of r.v.X; Answer: Step 1 We know that, total probability should be 1 for probability distributions. So 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛|1𝑥=0 𝑓(𝑥)𝑑𝑥 = 1 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛|10 𝑐𝑥𝑑𝑥 = 1 =( 𝑐𝑥 2 2 𝑐 1 ) =1 0 ⇒( )=1 2 𝑐=2 Step 2 𝑃(𝑋 < 0.5) = 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛|0.5 0 𝑓(𝑥)𝑑𝑥 = 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛|0.5 0 2𝑥𝑑𝑥 = (𝑥2)0.5 0 = (0.5)2 − 0 = 0.25 Step 3 C.d.f. is given by, 𝐹(𝑋) = 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛|0𝑥 𝑓(𝑥)𝑑𝑥 = 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛|0𝑥 2𝑥𝑑𝑥 = (𝑥 2 )0𝑥 = 𝑥2 C.d.f 𝐹(𝑋) = 𝑥 2 Question #20 Three cards are drawn without replacement from the 12 face cards (jacks, queens, and kings) of an ordinary deck of 52 playing cards. Let X be the number of kings selected and Y the number of jacks. a) Find the joint probability distribution of X and Y; b) Find the marginal probability distributions of X and Y. c) 𝑃[(𝑋, 𝑌) ∈ 𝐴] , where A is the region given by {(𝑥, 𝑦) | 𝑥 + 𝑦 ≥ 2} . Answer: Step 1 Total number of cards = 52 Total number of face cards = 12 In a pack of 52 cards, there are total 4 Kings and 4 Jacks. It is given that 3 cards are drawn. So, the possible values of X and Y are: 0, 1, 2 and 3. Now, the probability of drawing x kings and y Jacks can be calculated using the formula. 4 4 4 ( )( )( ) x y 3− x − y P ( X = x, Y = y ) = 12 ( ) 3 After drawing x kings, y jacks, (3 - x - y) would be left. Here, (x + y) should be always less than equal to 3. Step 2 a) The joint probability distribution of X and Y: Y/x 0 1 2 3 Total Y 0 1/ 55 6 / 55 6 / 55 1/ 55 14 / 55 1 6 / 55 16 / 55 6 / 55 0 28 / 55 2 6 / 55 6 / 55 0 0 12 / 55 3 1/ 55 0 0 0 1/ 55 Total X 14 / 55 28 / 55 12 / 55 1/ 55 1 Step 3 b) The marginal Probability distribution of X and y are Marginal Distribution of X X 0 1 2 3 Marginal Distribution 14 / 55 28 / 55 12 / 55 1/ 55 Marginal distribution of Y Y 0 1 2 3 Marginal Distribution 14 / 55 28 / 55 12 / 55 1/ 55 Step 4 c) To find: 𝑃[𝑋, 𝑌) ∈ 𝐴] where A is the region given by {(𝑥, 𝑦)|𝑥 + 𝑦 ≥ 2} That is 𝑃(𝑋, 𝑌 ∈ 𝐴) = 𝑃(𝑥 + 𝑦 ≥ 2) Now, 𝑃 = 1 − 𝑃(𝑥 + 𝑦 ≤ 1) = 1 − [[ P( X = 0, Y = 0] + [ P ( X = 0, Y = 1] + [ P ( X = 1, Y = 0]] 1 6 6 = 1−[ + + ] 55 55 55 13 42 = 1− = 55 55