7. In problem 6, suppose that another student decided to guess on all of the problems on the test. Suppose that there were 60 questions.

a) What would be the expected number of questions which would be correctly answered by guessing?

b) What is the standard deviation?

c) What is the probability that the student could get at least 20 correct by simply guessing on all of the quesions?

a) The expected number of correct guesses is

np = 60(1/5) = 12

A student could be expected to get around 12 questions correct by simply guessing. What is the probability that a student could pass the test by simply guessing? We first need to find the standard deviation.

b)

c) Rather than answer the question of what is the probability of passing the test, let us answer the question of what is the probability of getting at least 20 questions correct. On most tests, getting 20 questions correct is nowhere near passing. HOwever, if we look at the normal approximation to the binomial distribution,

we see that the probability of being that far from the expected number of correct guesses is pretty small. To find the probability, first compute the z score.

x is the left endpoint of the region marked in red. Since that region includes the entire box over the 20, its left endpoint is 19.5.

We look up the area which corresponds to a z-score of 2.42 in Table A,

The area in Table A is the area to the left of the z-score. However, we want the area to the right of the z-score so we must subtract the value in Table A from 1.

1 - 0.9922

= 0.0078

And there is less than 1% chance of getting that many questions correct by simply giessing. As you can see from the graph most of the area in red is over the 20. The chances of getting much more than 20 correct answers is extremely small.