STATISTICS PRE-TERM TEST (SOLUTIONS) INSTRUCTIONS 1. This test has 10 multiple choice questions 2. Each question has only one correct answer 3. This test is not timed. You can take as long as you like to complete the test. The boxplots show monthly sales revenue figures ($ thousands) for an office supplies company with locations in three different regions of the US (Northeast, Southeast and West).
1. Which region has the highest median sales revenue? A. B. C. D.
Northeast Southeast West All the three regions have equal median sales revenues
Answer: (A)
The horizontal bar inside the boxplot denotes the median. From visual inspection it is clear that it is highest for the Northwest region, which is around 145 units. 2. Which of the following statements are definitely TRUE about the variance of sales revenue? A. B. C. D.
West has a higher variance than Southeast Southeast has a higher variance than Northeast Northeast has a lower variance than West None of the above
Answer: (D)
Boxplot only provides information on the prominent quantiles and the outliers. It does not provide direct evidence on the variance of a random variable.
The number of intercity shipment orders arriving daily (in ‘000s) at a transportation company is a random variable X with the following probability distribution: x 0 1 2 3 4 5
P(X=x) 0.1 0.2 0.4 0.1 0.1 0.1
3. The company has a capacity to handle 3000 shipments per day during the regular working hours. If the demand exceeds the capacity, it has to pay overtime to its workers. If the unit overtime cost is $0.1/shipment, what is the expected daily overtime cost? A. B. C. D.
$30 $100 $200 $300
Answer: (A)
There are two outcomes of the random variable for which demand exceeds capacity, which is 4000 and 5000. So, the expected value of the excess demand is given by (4000-3000)*0.1 + (5000 – 3000)*0.1 units = 300 units. Then, the expected daily overtime cost is $ 300*0.1 = $30. 4. What is the probability that the number of shipments on a given day will exceed one standard deviation above the mean? A. B. C. D.
0 0.1 0.2 0.3
Answer: (C)
First, we use the formulas for calculating the expected value and the variance of X and obtain E(X) = 2200 units and Var(X) = 1960000. Thus, SD(X) = √ Var(X) = 1400 units. Then, the required probability is given by P(X>2200+1400) = P(X>3600). On inspecting the probability distribution, we obtain this as P(X=4000)+P(X=5000) = 0.1+0.1=0.2 5. X and Y are two independent random variables that have mean m and standard deviation s. Construct another random variable W = X + s. Which of the following statements is TRUE? A. Var (X+Y) = Var (W)
B. Var (X+Y) > Var (W) C. Var (X+Y) < Var (W) D. Var (W) = 0 Answer: (B)
Var(X+Y) = Var(X)+Var(Y) =2s2. Var (W) = Var(X+s) = Var(X)+Var(s) = s 2 because s is a constant. So , here Var(X+Y)> Var(W) 6. The time required for servicing transmissions is normally distributed with μ = 45 min and σ = 8 min. The service manager plans to have work begin on the transmission of a customer’s car 10 min after the car is dropped off and the customer is told that the car will be ready within 1 hour from drop-off. What is the probability that the service manager cannot meet his commitment? A. B. C. D.
0.3875 0.2676 0.5 0.6987
Answer: (B)
Let X denote the time required for servicing a car. Then, X~N(45,64). We want to find out P(X+10≥60) = P (X≥50). We calculate the z-value corresponding to 50, which is (50-45)/8 = 0.625. Then P (X≥50) = P(Z≥0.625) = 0.265986 from the Z-table. 7. Let X ~ N(100, 202). Find two values, a and b, symmetric about the mean, such that the probability of the random variable taking a value between them is 0.99. A. B. C. D.
94.1, 105.9 80.2, 119.8 48.5, 151.5 90.1, 109.9
Answer: (C) ≤ ≤
≤
≥
Since P(aa and X bb)= we havearound that P(X b) ≤=a)1-= 0.99 Further, because are0.99, symmetric thea)+P(X mean P(X P(X≥=b)0.01. = 0.01/2 = 0.005. We use the Z-table to calculate the z-value such that P(Z ≤z) = 0.005. From this, we get z=-2.575. Then, using the definition of z-value, z=(a-100)/20, we get a = 48.5. Because a and b are symmetric around the mean, b = 100+(100-48.5) = 151.5. 8. You are making a day-trip to a tourist island that has many attractions but no ATM and an entirely cash based economy. Based on your past trips to such exotic places, you think that your daily expenses are normally distributed with mean $5000 and standard deviation $350. What is the minimum amount of cash you should carry with you to keep the probability of running out of money to 2%? A. $5719 B. $4705
C. $5815 D. $4551
Answer: (A)
Let X denote the daily expenses. We know that X~N(5000,350). Also let m be the minimum amount to carry in hand no shortage appears., Then P(X ≤m) = 0.98. Using the z-tables, we get that the z-value corresponding to m is 2.054. Then, using the definition of z-value, we get (m-5000)/350 = 2.054. Thus, m = 5719. 9. A restaurant has three independent sources of revenue: eat-in orders, takeout orders and the bar. The daily revenue from each source is normally distributed with mean and standard deviation shown in the table below (all in $). Eat in Takeout Bar
Mean 5780 641 712
Std. Dev. 142 78 72
What is the probability that the total revenue will exceed $7,000 on a particular day? A. B. C. D.
0.5 0.23 0.77 0.68
Answer: (C)
Let X1, X2 and X3 be the three normal random variables denoting daily revenue from Eat in, Take out and Bar respectively. Here we need to find P(X1+X2+X3>7000). Let Y = X1+X2+X3. Then, E(Y) =E( X1)+E(X2)+E(X3) = 5780+641+712 = $7133, and Var(Y) = Var( X1)+Var(X2)+Var(X3) = 142 2+782+722 = 31432. Thus, SD(Y) = √31432 = $177.29. We use this information to calculate the z-value corresponding to 7000 as follows: z = (7000-7133)/177.29 = -0.75. Then, P(Y>7000) = P(Z> -0.75) = 77% from the Z-tables. 10. An energy startup company founded by a bunch of MBAs is designing a new generation solar power generator. Tests show that the output of the generator is normally distributed with mean 15.6KW. What should be the target standard deviation if the designers want the probability of meeting the peak load of 20KW to be 20%? A. B. C. D.
4.4 8.66 8.73 5.26
Answer: (D)
Let X denote the output of the generator in KW. Then, we know that X~N(15.6, σ2). We are also given that P(X≥20) = 0.2. Using the z-table, we know that z-value corresponding to this probability is 0.8365. Now, using the definition of z-value, 0.8365 = (20-15.6)/σ. Solving this equation, we get that σ= 5.26.
