Question

In: Statistics and Probability

I want to use R markdown to do the following questions and render a pdf for...

I want to use R markdown to do the following questions and render a pdf for all the answers!!!

Q1.

Suppose we toss 4 coins (each having heads probability = (1/2). Let X denote the random variable: (number of heads) - (number of tails).

(a) What is the range of X? (give exact upper and lower bounds along with a line of explanation)

(b) What is the probability mass function of

(c) What is the cumulative density function of X

Q2. Using R, simulate tossing 4 coins as above, and compute the random variable X defined in problem 1. Estimate the probability mass function you computed in part (b) of problem 1 by simulating 1000 times and averaging.

Solutions

Expert Solution

X = Number of heads - number of tails

If we toss 4 coins, then possible number of heads we can get : 0,1,2,3,4

Similarly, possible number of tails we can get : 0,1,2,3,4

Then the random variable can take the values :

-4,-2,0,2,4

(The values listed above are calculated in the following way:

if we get 4 tails and no heads, then X = 0-4 = -4 , if we get 3 tails and 1 head, X = 1-3 = -2 , 2 heads and 2 tails gives us X = 2- 2 =0 and so on)

Hence, upper bound of X is 4 (when we get 4 heads and no tails) and lower bound is -4 (when we get 4 tails and no heads)

Hence range of the random variable X is (Upper bound - Lower bound) = (4 - (-4) ) = 8

b)

X can take value -4 when all the outcomes are tail , and probability of this event is

= 0.0625

X can take value +4 when all the outcomes are head, probability of this event is

=0.0625 ( as we are considering an unbiased coin where probability of getting a head and probability of getting a tail is same and is 0.5)

Now , X can take value -2 when there are 3 tails and 1 head. Here we can consider a binomial distribution with n=4 and probability of success = 0.5 where success is considered as getting a head when a coin is flipped.

Hence, we know the pmf of a binomial distribution with n= 4 and p = 0.5 is

Hence for this event, our required probability is

= 0.25

where the random variable Y denotes the number of heads.

also,

random variable X takes the value 0 when we get 2 heads and 2 tail and probability of this event is

Again, random variable X takes the value 2 when there are 3 heads and 1 tail and probability of this event is

Hence, the probability distribution of the random variable X is given by

Cumulative Density Function of X


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