Question

1.)

In the previous question (the one about the multiple-choice quiz), the random variable X is binomial with parameters:

• n = 1/5, p = 15
• n = 15, p = 1/2
• n = 15, p = 1/5
• n = 15, p = 0

2.)

Assume that a procedure yields a binomial distribution with a trial repeated n=5n=5 times. Use technology to find the probability distribution given the probability p=0.164p=0.164 of success on a single trial.

(Report answers accurate to 4 decimal places.)

k	P(X = k)
0
1
2
3
4
5

3.)

Suppose that 65% of all voters prefer Candidate A.

If 3 people are chosen at random for a poll, what is the probability that exactly 1 of them favor Candidate A?

4.)

Using the Binomial distribution,

If n=10 and p=0.2, find P(x=3)

5.)

The probability that you will win a game is 0.640.64.

If you play the game 588 times, what is the most likely number of wins?
(Round answer to one decimal place.)
μμ =

Let XX represent the number of games (out of 588) that you win. Find the standard deviation for the probability distribution of XX.
(Round answer to two decimal places.)
σ =

The minimum usual value for a random variable is μ-2.5σ and the maximum usual value is μ+2.5σ. You already found μμ and σσ for the random variable XX.


Find the usual range of XX values. Make the calculations using the exact values of μ and σ (not the rounded values you entered above). Enter final answer as an interval using square-brackets and only whole number values. To get the correct whole number values, round your lowest value up to the nearest whole number and round your highest value down to the nearest whole number.


Usual values =

Answer 1

Please don't hesitate to give a thumbs for the answer incase you' re satisfied with the answer.

2.

We have been given p = .164

We will use the binomial distribution pdf function to solve this:

For k = 0 , the P(x=k) = 5C0*(.164)^0 *(1-.164)^5 = 0.4083
For k = 1 , the P(x=k) = 5C1*(.164)^1 *(1-.164)^4 = 0.4005
For k = 2 , the P(x=k) = 5C2*(.164)^2 *(1-.164)^3 = 0.1571
For k = 3 , the P(x=k) = 5C3*(.164)^3 *(1-.164)^2 = 0.0308
For k = 4 , the P(x=k) = 5C4*(.164)^4 *(1-.164)^1 = 0.0030
For k = 5 , the P(x=k) = 5C5*(.164)^5 *(1-.164)^0 = 0.0001