(1) Call the -sided, -sided, and -sided dice, dice respectively. Let be the event that dice rolls a .
Then , , and .
(2) Let be the number of ‘s that are rolled. Then . Thus,
by linearity of expectation.
Thus, we expect to see ‘s if we roll these dice together.
02
(1) Let be the value of the first coin drawn, let be the value of the second coin drawn, and let be the value of the third coin drawn. The central trick to efficiently solve this problem is to notice that are all identically distributed.
One can see this by the following argument: using an ordered triple , write down all possible permutations of drawings. Notice that the number of triples where is a dime is equal to the number of triples where is a dime is equal to the number of triples where is a dime.
We can further extend this observation to all the values. Thus, the distributions of are all the same.
(2) Another, nicer, argument is to notice that we can swap and in these ordered triples without changing the overall set, and similarly for and there exists a bijection between ; ; and identical distribution.
Thus, we have that .
(3) Let be the sum of the values of the three coins. Then,
Now,
Thus, .
03
(1) Let be a sequence of indicators where is the event that the -th entry differs from the -th entry. By a similar argument to above, the are identically distributed for each , and are, in fact, independent.
(2) Let be the number of pairs of entries that differ from each other. Then . By the above,
(3) Let denote the -th entry. Now,
Thus, .
04
(1) The normal approximation in this case is applicable since:
Assumptions:
Frank eats a large number of hot dogs the sample size, or , is sufficiently large
We assume that the amount of time Frank spends on each hot dog does not depend on how many he has had previously the times to consume each hot dog are independent and identically distributed
(2) Let be the time taken to eat the -th hot dog. Let be the time taken to eat hot dogs. Then seconds with seconds.
Since minutes is seconds, by the CLT we have:
05
06
In this case the normal approximation is applicable since we have a large sample size (need a large number of filters to last hours) and they follow follow independent but identical distributions.
Let be how long the -th filter lasts. Let where we want to find such that . By normal approximation and the Central Limit Theorem, we have
and thus filters are required.
07
Let and be the number of heads and tails respectively. Then we have the following two conditions:
Thus, .
Let be the event that the -th flip is a head. Then for each and .
Thus, by CLT, .
By the normal approximation, using the continuity correction, we have:
08
(a)
Let be the indicator that a man was chosen in the -th period. Then for each , and the are independent for each . Let be the total number of times a man was chosen. We can use a similar argument to Problems 2 or 3, or we can simply use linearity of expectation:
(b)
(1) Using the standard formula for the variance of a sum of random variables, we have:
Since the variables are identically distributed, their variances are equal. Thus,
(2) Now, the sum has terms, and since each is identically distributed, each term is identical. Thus,
for some fixed .
(3) We then have
Finally, , and thus
Plugging these values in, we have .
09
(a)
(b)
(c)
The standardized variable is:
Then:
10
Let .
From the CDF table, :
11
Sample mean RV:
has and .
Standardize this and approximate with a normal distribution: