PROBLEM 1. A first list of 100 numbers contains 20% ones and 80% threes (in some

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PROBLEM 1.

A first list of 100 numbers contains 20% ones and 80% threes (in some order). A second list of 100 numbers contains 50% twos and 50% fives (in some order). A third list is obtained by adding the first number from the first list to the first number from the second list to get the first number in the third list, etc., so that the k-th number in the third list is the sum of the k-th numbers in the first two lists.

(a) What is the average of the 100 numbers in the third list?

(b) Suppose that instead you get the k-th number in the third list by multiplying the k-th numbers in the first two lists. What is the average of the numbers in the third list? Or can't you tell what that value is?

PROBLEM 2. Last semester I had 40 students in my class. The average score on the first midterm was 20.

(a). Assuming that no student got a negative score, what is the largest possible number of students who got a score of 50 or more?

(b). If the variance of the scores (= average squared difference from mean score) was 25, could there have been 10 scores of 30 or more?

PROBLEM 3. Exercise 3.3.8, except with respective probabilities 1/6 , 1/5, and 1/4.

PROBLEM 4. Exercise 3.3.3, but with mean 3 and variance 1, and also

(e) Var (X-Y)

PROBLEM 5. There are 10 tickets in a box, each with a number on it. The average of those 10 numbers is zero. The average value of the squares of those numbers is 5.

(a). If you pick two tickets at random, WITH replacement, what is the average product of the numbers drawn?

(b). If you pick two tickets at random, WITHOUT replacement, what is the average product of the numbers drawn?

PROBLEM 6

Let F_n be the fraction of tosses that come up heads in n tosses of a fair coin.

(a), Using the normal approximation, about how big does n have to be so that P{ | F_n -1/2 | > .01 } < .01 ?

(b) How big does n have to be so that the Chebyshev inequality implies P{ | F_n -1/2 | > .01 } < .01 ?

CHALLENGE PROBLEM: Seven identical 50-foot garden hoses are in a big tangle on my lawn, with none of the ends connected. Recall that each garden hose has a "male" end and a "female" end, and that ends can be connected only if they are of different types.

Suppose the male ends are randomly paired and connected to the female ends. Then the hoses will form a random number L of loops, at least 1 and at most 7. There will be 7 loops if each male end happens to get connected to the female end of the same hose. There are lots of ways to end up with one big loop.

(a). Find P( L= 7).

(b). Find P( L= 1)

(c). Find E(L).

(d) Find var(L)

Explanation / Answer

Problem 1:

First list contains 20% of ones and 80% of threes

Therefore average value of the sum of the numbers = 0.2*1 + 0.8*3 =2.6

The second list contains 50% twos and 50% five's

Therefore the average value of sum of numbers in the second list is = 0.5*2 + 0.5*5 = 3.5

The average value of sum of numbers in the third list is = (2.6+3.5)/2 =3.05

There are total 100 numbers in the third list.

The average of 100 numbers = 3.05*100/100 =3.05

b) If k th number isobtained by multiplying kth number in the first list and kth number in the second list

Then the average number is = 2.6*3.5 =9.1

The average of numbers in the third list is 9.1

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