Stat230-Midterm_2023W1 - Solution

STAT 230 Midterm Exam 19 October 2023 INSTRUCTOR: Yas YAMIN Irving K. Barber School of Arts and Sciences UBC, Okanagan Campus This test has 4 questions, for a total of 30 points, and you have 75 min to answer the questions. READ THE QUESTIONS CAREFULLY In order to get full credit you must SHOW YOUR WORK Answer the questions in the spaces provided on the question sheets. Non-graphing calculators are allowed. No textbook, laptops, cell phones, or other electronic devices are permitted. Last Name, First Name (print). Student Number Signature: Question Points Score 1 10 2 4 3 7 4 9 Total: 30 This exam has 5 pages including this cover page

STAT 230 Midterm Exam (continued) 1. Answer the following questions; (a) 2 Suppose the number of tornadoes to occur in a certain region follows a Poisson distribution with an average of 2 per year. What is the probability that less than 3 tornadoes will occur in this region in any 2 year period? Solution: Let Y be the number of tornadoes. rate = 4 for any 2 year period P ( Y < 3) = P ( Y = 0) + P ( Y = 1) + P ( Y = 2) = 0 . 0183 + 0 . 0732 + 0 . 1465 = 0 . 238 (b) 1 What is the number of possible arrangements of the letters in the word ”distributions” ? Solution: 13! 3!2!2! = 2594549200 (c) 2 United Airlines Flight 433 is a nonstop flight from Boston’s Logan airport to San Francisco International Airport. An important factor in scheduling such flights is the actual airborne flying time from takeo ↵ to touchdown. Suppose the airborne flying time is distributed as normal with mean 370 minutes and standard deviation 16 minutes. What is the probability that the airborne flying time is more than 383 minutes in one day? Solution: z = 383 - 370 16 = 0 . 8125 p ( Z > 0 . 8125) = 1 - p ( Z < 0 . 8125) = 0 . 209 (d) 2 The standard deviation for a data set is 10. If each observation in the data set is multiplied by 3, calculate the variance of the resulting data set. Solution: var ( ax ) = a 2 var ( x ) var ( x ) = 100 var (3 x ) = (3) 2 var ( x ) = 900 (e) 2 In a certain population, 30% are smokers. If a random sample of 60 individuals is selected from the population, find the expected number of smokers (E[S]) and the variance of the number of smokers (V(S)) in the sample. Solution: S is a binomial random variable with parameters n=60 and p=0.3. E[S] = np = 60(0.3) = 18 V(S) = np(1-p) = 60(0.3)(0.7) = 12.6 Page 2 of 5

STAT 230 Midterm Exam (continued) 3. The number of car accident fatalities in any one week period is a random variable X having probability distribution P(X= x )= p(x) where x 2 3 4 5 p(x) 0.15 0.35 0.2 (a) 1 Fill in the blank in the table above. (b) 2 Find P(X < 4). (c) 4 Find the standard deviation of X. Solution: (a). 1 - 0 . 15 - 0 . 35 - 0 . 2 = 0 . 3 (b). P ( X < 4) = p (2) + p (3) = 0 . 15 + 0 . 35 = 0 . 5 (c). V ( X ) = E [ X 2 ] - E [ X ] 2 E [ X ] = 2(0 . 15) + 3(0 . 35) + 4(0 . 3) + 5(0 . 2) = 3 . 55 E [ X 2 ] = 4(0 . 15) + 9(0 . 35) + 16(0 . 3) + 25(0 . 2) = 13 . 55 V ( X ) = 13 . 55 - (3 . 55) 2 = 0 . 9475 SD ( X ) = p 0 . 9475 = 0 . 9734 Page 4 of 5

STAT 230 Midterm Exam (continued) 4. The probability density function for a certain type of measurement Y is given by f ( y ) = k y 2 , if 1  y  2 and f ( y ) = 0, otherwise. (a) 3 Determine the value of k . Solution: Z 1 -1 k y 2 dy = 1 Z 2 1 k y 2 dy = 1 - k y | 2 1 = 1 k 2 = 1 = ) k = 2 (b) 2 Find E [ Y 3 ]. Solution: E [ Y 3 ] = Z 1 -1 y 3 2 y 2 dy = Z 2 1 2 ydy = y 2 | 2 1 = 3 (c) 4 Find F ( y ), and use this result to find P (1 . 5 < Y  2) Solution: F ( y ) = Z y -1 2 y 2 dy = Z 2 1 . 5 2 y - 2 dy = - 2 y P (1 . 5 < Y  2) = F (2) - F (1 . 5) = - 1 - ( - 2 1 . 5 ) = 1 3 Page 5 of 5 y ≤ I 1 Pt " = % " ≤ " " " I 472 I pt 1 Pt