Statistics FOR WIZARD KIM

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Week 4 Discussion Example:

A high school student taking the state standardized achievement test, received blank test booklet for

one section of the test. He decided to go ahead and fill in the True/False answer sheet anyway. Since

there were only two possible answers to each question he figured that he could get at least 50% on the

section.

Response parameters:

What do you think about the student’s idea? Why?

Aside from being a very dumb idea. He has confused the probability of guessing the correct

answer on a single problem with the score he would receive for the entire test.

What do you think his chances of getting at least 20% on the section are?

If we assume there will be 50 questions on the section, my friend would need to guess correctly

on 25 or more questions. We want to find the probability of getting 25 or more correct answers

out of 50 questions. This is a Binomial Distribution problem with the probability of success p =

0.50 and n = 50, and X = the number of correct answers. We want to find P(X ≥ 25)

We could find P(X = 25) + P(X = 26) + P(X = 27) + ••• + P(X = 50), but since we know the

probability that X is between 0 and 50 is equal to 1, we can find the compliment P(X < 25) and

subtract it from 1.

P(X ≥ 25) = 1 – P(X ≤ 24).

If we wanted to use a Table of Binomial Probability (p = 0.50, n = 50) we would sum P(X = 0) +

P(X = 1) + P(X = 2) + P(X = 3) + ••• + P(X = 24) and subtract that from 1, But we can use the Excel

function BINOM.DIST(24, 50, 0.50, True) to find P(X ≤ 24)

BINOM.DIST(4, 20, 0.25, True) = 0.4439

P(X ≥ 5) = 1 – 0.4148 = 0.5561

He has a 55.61% chance of getting at least a 50% on a 50 a question test.

Do the number of questions in the section make a difference? If it does, should he hope there are 50 or

200 questions in the section

If there were 200 questions on the exam my friend would need to get 100 or more questions

correct. Using the same approach as above.

P(X ≥ 100) = 1 – P(X ≤ 99).

P(X ≤ 99) = BINOM.DIST(99, 200, 0.50, True) = 0.4718

P(X ≥ 40) = 1 – 0.4718 = 0.5282

He would have a 52.82% chance of getting at least a 50% on a 200 a question test.

He should hope there are 50 question rather than 200 question in the section.

(Tip: It may help if you create a table of Binary probabilities with p = 0.20 and n = number of questions

on the exam. Also, don’t confuse the probability of getting exactly 20% of the questions correct and

getting at least 20% of the questions correct)