See attached question

Question

Trevor is interested in purchasing the local hardware/sporting goods store in the small town of Dove Creek, Montana. After examining accounting records for the past several years, he found that the store has been grossing over $850 per day about 50% of the business days it is open. Estimate the probability that the store will gross over $850 for the following. (Round your answers to three decimal places.)

(a) at least 3 out of 5 business days

 

(b) at least 6 out of 10 business days

(c) fewer than 5 out of 10 business days

(d) fewer than 6 out of the next 20 business days

If the outcome described in part (d) actually occurred, might it shake your confidence in the statement p = 0.50? Might it make you suspect that p is less than 0.50? Explain.

Yes. This is unlikely to happen if the true value of p is 0.50.Yes. This is likely to happen if the true value of p is 0.50.    No. This is unlikely to happen if the true value of p is 0.50.No. This is likely to happen if the true value of p is 0.50.

(e) more than 17 out of the next 20 business days

________________________

If the outcome described in part (e) actually occurred, might you suspect that p is greater than 0.50? Explain.

Yes. This is unlikely to happen if the true value of p is 0.50.Yes. This is likely to happen if the true value of p is 0.50.    No. This is unlikely to happen if the true value of p is 0.50.No. This is likely to happen if the true value of p is 0.50.

question 11}

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 55% of all computer programmers are introverts. (Round your answers to three decimal places.)

(a) At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

 

What is the probability that 5 or more are extroverts?

 

What is the probability that all are extroverts?

 

(b) In a group of 4 computer programmers, what is the probability that none are introverts?

 

What is the probability that 2 or more are introverts?

 

What is the probability that all are introverts?

Answer 1

Preity,

The problem at hand here is to determine the probability of one or more successful outcomes out of a given number of trials, and given a probability of one success happening out of one individual trial. These exercises follow the binomial distribution (see more below).

So let's look at your first example:

We are looking at 5 business days (number of trials), and wondering if the store will gross more than $850 (defined as a success) for at least 3 days (number of successes), given that 50% of the time the store does (probability of an individual success).

Each day there are two possible outcomes (success/failure, or 1/0); so after two days you will have 4 possibilities (1,1); (1,0); (0,0); (0,1); after three days you will have 8 possibilities, 16 after four days, and ultimately 32 possible combinations after 5 days (0,0,0,0,0), (0,0,0,0,1) all the way to (1,1,1,1,1).

The total number of combinations possible can be written as 2^5 = 32.

Since we want to count the number of combinations where we have at least 3 successes (for example: 1,1,0,1,0), we will count the number of combinations where we have 3, 4 and 5 successes.

If you write all possible combinations, you will see there are:

1. one way to have 5 successes (1,1,1,1,1)
2. five ways to have 4 successes (0,1,1,1,1);(1,0,1,1,1);(1,1,0,1,1); (1,1,1,0,1); (1,1,1,1,0)
3. ten ways to have 3 successes (1,1,1,0,0); (1,1,0,0,1); etc.

So the number of ways to combine three or more successes out of five trials is 1+5+10 = 16

Since the probability of an individual event is 50%, the probability of having three or more successes out of five trials is therefore the number of possible successes divided by the number of possible outcomes, i.,e. 16/32 = 0.50. In math, you would write this as [C(5,5)+C(5,4)+C(5,3)]/2^5 = 0.50

In Excel you can use the COMBIN() function:

(COMBIN(5,5)+COMBIN(5,4)+COMBIN(5,3))/2^5 = 0.5

The C(n,k) or COMBIN(n,k) works well for problems where there is equal probability of success and failure, i.e. p=0.50. For different probabilities of success, you must refer to the binomial distribution.

The probability P(k) of k successes in n events for a binomial distribution with probability of success p can be expressed as: P(k) = p^k*(1-p)^(n-k)*C(n,k)

Fortunately, you can use (hopefully) software such as Excel to solve this using the BINOM.DIST() function:

P(k) = BINOM.DIST(k,n,p,FALSE) returns the probability of k successes;

and

CDF = SUM(0 to k) of (P(k)) = BINOM.DIST(k,n,p,TRUE) returns the sum of all probabilities of up to k successes out of n trials, or the cumulative probability for k successes.

in our case:

P(>=3,5) =1-P(<=2,5) = 1-BINOM.DIST(3,5,0.5,1) = 0.5

Similarly, you get:

P(>=6,10) = 1-P(<=5,10) = 1-BINOM.DIST(5,10,0.5,1) = 0.377

P(<5,10) = P(<=4,10) = BINOM.DIST(4,10,0.5,1) = 0.377 (same as above! - check out Pascal's triangle to learn why!)

P(<6,20) = P(<=5,20) = BINOM.DIST(5,20,0.5,1) = 0.021

Now, you realize there is about 2% chance of five successes (store making more than $850) out of twenty days. If, indeed, you were to see that happening, you should suspect that the initial probability of making more than $850 per day is less than 50%; a typical threshold for suspecting foul play is 0.05 (this is called the significance level).

P(>17,20) = 1-P(<=17,20) = 1-BINOM.DIST(17,20,0.5,1) = 0.000

This means there is practically no chance you will make more than $850 for more than 17 days out of 20, if the probability of an individual success is 50%. You should suspect the probability of making more than $850 per day to be higher than 50%.

Your second set of problems can be solved in a similar fashion, adjusting the initial probabilities.

Hope this helps!

Yves