1. True or False? The standard deviation \sigma cannot be a negative number.
Solution TRUE. The definition of the standard deviation is the positive square root of a sum of squares, so we are taking the positive square root of a non-negative number, this cannot be neagtive.
2. Suppose that the weight of adult males in the US is normally distributed with mean 179.6 and standard deviation 31.2. What approximate percentage of adult males weigh less than 150?
Solution Since 179.6-150-29.6 we are 29.6/31.2=0.95 standard deviations away from the mean. if we look up z=0.95 the area is about 0.33, so the remaining area s 0.17. We have about 17% weighing less than 150.
3. A 100 question multiple choice test is given. 60 of the questions have 5 choices, and the remaining 40 questions have 4 choices A student does not read the questions, instead she randomly chooses an answer. How many correct answers can she expect to get?
Solution 60/5+40/4=12+10=22.
4. Find the mean and the standard deviation of the following set of 5 numbers: {x_1,x_2,x_3,x_4,x_5}, where x_i=\sum_{j=1}^i j.
Solution x_1=\sum_{j=1}^1 j=1, x_2=\sum_{j=1}^2 j=1+2=3, x_3=\sum_{j=1}^3 j=1+2+3=6, x_4=\sum_{j=1}^4 j=1+2+3+4=10, x_5=\sum_{j=1}^5 j=1+2+3+4+5=15. The mean is (1+3+6+10+15)/5=7, the st dev is
\sqrt((36+16+1+9+64)/5)=\sqrt(126/5)=5.0 about.
5. Explain why the following is true: \sum_{k=1}^n (n choose k)=2^n-1.
Solution If we put x=y=1 into the binomial theorem we have
2^n=\sum_{k=0}^n (n choose k)=(n choose 0)+(n choose 1)+(n choose 2)+...+(n choose n).
This problem has exactly this sum but without the (n choose 0)=1 term. So we just subtract this term from both sides.
1. Find the standard deviation and mean of {11,0,-11,8}. Leave your results as fractions, not decimals.
Solution mean=(11+0-11+8)/4=2, st dev = \sqrt((81+4+169+36)/4)= \sqrt(145/2).
2. True or False? Give your reason. The mean of a list of data must be no greater than the largest number on the list.
Solution True. we are taking the average, and the average cannot be greater than the largest number on the list.
3. On the SAT math exam the mean is 512 with standard deviation 78. If my score is 666, what approximate percentage of the examinees are within one standard deviation of my score?
Solution 666-512=154 so we are 154/78=1.97 st dev from the mean. So we need the area of th enormal curve between 0.97 and 2.97 st deviations, this is from table 11.3 about 0.499-0.34= about 16%.
4. A fair coin is flipped 3 times. If 0 heads appears, you get $1, if 1 head appears you get $2, if 2 heads appear you get $4, if 3 heads appear you get $8. What are your expected winnings?
Solution
0 heads: 1/8*1=1/8
1 head 3/8*2=6/8
2 heads 3/8*4=12/8
3 heads 1/8*8=8/8
for a total of 27/8 dollars.
5. Write in summation notation the following theorem: The sum of the first n positive odd integers is the square of n.
Solution \sum_{k=1}^n (2k-1) =n^2.
1. True or False? Give your reason. For any set of data, the average of the square must be greater than or equal to the square of the average.
Solution TRUE. We know from problem 11.3.7 that the standard deviation squared is exactly the average of the square minus the square of the average, and this definitely is a positive (or zero) number since it is a square.
2. In a dart game you have a 10% chance of a bulleye. Find the approximate probability that in 1000 throws you will have between 90 and 120 bulleyes.
Solution With n=1000 and p=1/10 the mean is n*p=100 and the st dev \sigma=\sqrt(1000*1/10*(9/10))=\sqrt(90)=9.5 about. So our interval is [-10.5/9.5,20.5/95]=[-1.1,2.16]. if we use Table 11.3, for 1.1 we get 0.364 and for 2.16 we get 0.484, so the answer is 0.364+0.484=0.848, about 85%.
3. Let a_i=(4 choose i), find \sum_{j=1}^3 a_j*a_{j+1}.
Solution a_0=1, a_1=4, a_2=6, a_3=4, a_4=1. We need
a_1*a_2+a_2*a_3+a_3*a_4=4*6+6*4+4*1=52.
4. Suppose we have a list of 100 numbers whose mean is 0 and whose standard deviation is 1. We now add the numbers {4,-4} to our list. Show that the mean of the new list is still zero, but the new standard deviation is greater than 1. Can you find the new standard deviation?
Solution Since our original average is 0, when we add all 100 numbers the answer is 0. So if we add to these 100 numbers our new numbers, 4 and -4, the sum of all 102 numbers will still be 0, so the new average is 0.
The st dev squared is 1. We know that the st dev is the average of the squares minus the average (=0) squared. So the sum of the squares of all 100 numbers must be be 100. When we add 4 and -4 to the list, the new sum of the squares is 100+16=16=132. The average square is now 132/102, so the new st dev is \sqrt(132/102) which larger than 1.
5. In a dice game I roll two dice. If I get a seven I win $5, if I get eleven I win $15, otherwise I pay $2. What are my expected winnings?
Solution
roll a 7: 1/6*5
roll an 11: 2/36*15
otherwise: 28/36*(-2)
the total is 5/6+30/36-56/36=-21/36 dollars. I lose money in this game.
1. Give an example of a set of data of size 4 whose mean is 0 and whose standard deviation is less than 1.
Solution Try {0,0,1,-1}. The mean is 0 and the st dev is \sqrt(1/2). Even {0,0,0,0} works.
2. A so-called psychic guesses colors of cards (red, blue, or yellow). What is the approximate probability that by randomly guessing he can guess at least 50 correct out of 100 tries?
Solution n=100, p/13, mean =33.33 st dev =\sqrt(100*1/3*2/3)= 4.74. We need the area beyond (49.5-33.3)/4.74=3.74=z st deviations. This will be a tiny number, z=3.74 it is not on our Table 11.3. It is about .499902, so the answer is .000092.
3. True or False. For the mean of a set of data to be zero, each number in the set must be paired with its negative which is also in the set.
Solution FALSE. {2,1,-3}.
4. You draw a single card from a 52 card deck. I now guess what the card is, for example I may guess queen of hearts. If my guess has the correct suit, you pay me $3, If my guess has the correct rank (queen here) you must pay me $2. So I could win $3, $2, or $5 from you. Otherwise, if my guess is all wrong I pay you $1. What are my expected winnings?
Solution
rank and suit correct: 1/52*5
suit correct, rank wrong 12/52*3
rank correct wrong suit 3/52*2
all wrong 36/52*(-1)
The total is 5/52+36/52+6/52-36/52=11/52 dollars.
5. True or False? (\sum_{i=1}^n x_i)^2= \sum_{i=1}^n x_i^2.
Solution FALSE (1+1)^2 does not equal 1^2+1^2.