STAT 200 QUIZ 2 Solutions Section 6380 Fall 2013 The quiz covers Chapters 4, 5 and 6. 1. (8 points) If the IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. (a) (3 pts) Find the probability that a randomly selected person has an IQ score between 88 and 112. (Show work) We want to find P(88 < x < 112). First, we have to convert the given normal distribution to standard normal distribution using the ( x µ ) formula z =, where μ =100 and = 15. z-score corresponding to 88 is (88 100) / 15 = -12 / 15 = -0.8 z-score corresponding to 112 is (112 100) / 15 = 12 / 15 = 0.8 Hence the required probability is equivalent to P(-0.8 < z < 0.8) as shaded area in the graph. Use Table A-2, we find -0.8 corresponds to an area of.2119, and 0.8 corresponds to an area of.7881. So the shaded area = 0.7881 0.2119 = 0.5762 1
(b) (2 pts) If 100 people are randomly selected, what is the standard deviation of the sample mean IQ score. (Show work) The sample mean is normally distributed with µ x = µ and x = n Hence the standard deviation of the sample mean IQ score is 15 15 x = = = = 1.5 n 100 10 (c) (3 pts) If 100 people are randomly selected, find the probability that their mean IQ score is greater than 103. (Show work) The sample mean is normally distributed with µ x = µ and x = n Hence the mean IQ score is a normal distribution with 15 15 µ x = µ = 100 and x = = = = 1.5 n 100 10 We want to find Px> ( 103). First, we have to convert the given normal distribution to standard normal distribution: z-score corresponding to 103 is (103 100) / 1.5 = 3 / 1.5 = 2 Hence the required probability is equivalent to P(z > 2) as shaded area in the graph. Use Table A-2, the area to the left of 2 is 0.9772, so the required probability is 1 0.9772 = 0.0228. 2
2. (8 points) Imagine you are in a game show. There are 4 prizes hidden on a game board with 16 spaces. One prize is worth $4000, another is worth $1500, and two are worth $1000. You have to pay $50 to the host if your choice is not correct. (a) (6 pts) What is your expected winning in this game? (Show work) The probabilities and payoffs are summarized in the following table : x P(x) x * P(x) $4000 1/16 $250.00 $1500 1/16 $93.75 $1000 2/16 $125.00 -$50 12/16 -$37.50 Total $431.25 The table shows that the expected winning is E=Σ [ xpx ( )] = $431.25 (b) (2 pts) If you are offered a sure prize of $400 in cash, and you can just take the money without playing the game. What would be your choice? Take the money and run, or play the game? Please explain your decision. If you are risk-averse, you will take the money and run since the expected winning is just a little more than the $ 400 in cash. However, statistically speaking, playing the games many, many times is better than the cash prize of $ 400 3. (7 points) Mimi just started her tennis class three weeks ago. On average, she is able to return 15% of her opponent s serves. If her opponent serves 10 times, please answer the following questions: (a) (5 pts) What is the probability that she returns at most 2 of the 10 serves from her opponent? (Show work) This is a binomial distribution with n = 10, p = 0.15 and q = 1 0.15 = 0.85 The required probability is Px ( 2) = Px ( = 0) + Px ( = 1) + Px ( = 2). 3
10! Px= = = (10 0)!0! 0 10 ( 0) (0.15) (0.85) 0.196874 10! Px= = = (10 1)!1! 1 9 ( 1) (0.15) (0.85) 0.347425 10! Px= = = (10 2)!2! 2 8 ( 2) (0.15) (0.85) 0.275897 Hence, the probability is 0.19684 + 0.347425 + 0.275897 = 0.820196 (b) (2 pts) How many serves can she expect to return? (Hint : What is the expected value?) (Show work) Use formula 5-6 on page 230 of the textbook, µ = np = (10)(0.15) = 1.5 Mimi can expect to return 1.5 serves. 4. (4 points) Men s heights are normally distributed with mean 69.0 inches and standard deviation 2.8 inches. Mimi is designing a plane with a height that allows 95% of the men to stand straight without bending in the plane. What is the minimum height of the plane? (Show work) The figure below shows the normal distribution with the height x that we want to identify. The shaded area represents the 95% of men can fit through the airplane that Mimi is designing. In Table A-2 we search for an area of 0.9500 in the body of the table. The area of 0.9500 is between the Table A-2 areas of 0.9495 and 0.9505, but there is an asterisk and footnote indicating that an area of 0.9500 corresponds to z = 1.645. We can solve for x by using Formula 6-2 4
x µ x 69 z = becomes 1.645 = 28 or x = µ + ( z ) = 69 + (1.645 28) = 73.606 Hence the minimum height of the airplane is 73.606 inches. 5. (6 points) There are 7 seniors and 3 juniors in the statistics club, and a team of 5 will be randomly selected to attend the Joint Statistical Meetings in Montréal. (a) What is the probability that all 3 juniors are picked in this team of 5? (Show work) 3! 7! There are ( 3 C 3)( 7 C 2) = 21 (3 3)!3! (7 2)!2! = combinations of choosing all 3 juniors and 2 seniors. Therefore, the probability is 21 / 252 = 1/12 = 0.08333 (b) What is the probability that none of the juniors are picked in this team of 5? (Show work) 3! 7! There are ( 3 C 0)( 7 C 5) = 21 (3 0)!0! (7 5)!5! = combinations of choosing 0 juniors and 5 seniors. Therefore, the probability is 21 / 252 = 1/12 = 0.08333 (c) What is the probability that 2 of the juniors are picked in this team of 5? (Show work) 3! 7! There are ( 3 C 2)( 7 C 3) = 105 (3 2)!2! (7 3)!3! = combinations of choosing 2 juniors and 3 seniors. Therefore, the probability is 105 / 252 = 5/12 = 0.41667 5
6. (4 points) There is a 1% delinquency rate for consumers with credit rating scores above 800. If UMUC Credit Union provides large loans to 10 people with credit rating scores above 800, what is the probability that at least one of them become delinquent? (Show work) Assume that x is the number of people becoming delinquent, then this is a binomial distribution with n = 10, p = 0.01 and q = 0.99. We want to find P(x > = 1). Use complement rule, 10! 0 10 P(x >= 1) = 1 P(x = 0) = 1 (0.01) (0.99) = 1 0.904382 = 0.095618 (10 0)!0! 7. (3 points) A telemarketing company has two customer service teams. Team A has 20 agents and Team B has 30 agents. 10 agents in Team A and 20 agents in Team B contact customers via e-mails, and the rest contact customers via phone. Find the probability of getting someone who is from Team A, given that the selected person uses phone to contact customers. (Show work) E-mails Phone Total Team A 10 10 20 Team B 20 10 30 Total 30 20 50 Let event A be the selected person is from Team A and event H be the person uses phone. We want to find conditional probability P(A H) = P(A and H) / P(H). P(A and H) = 10/50 P(H) = 20/50 Hence, the conditional probability is 10 / 50 = 1 = 0.5 20 / 50 2 6