3. Linda took a multiple choice test. She did pretty well, but there were 5 questions where she had to guess. Each question had 4 choices for answers. Make up a probability distribution for the number of correct guesses. Find the mean and standard deviatiion. Draw the histogram. What is the probability that she got at least 2 right?

We use the formula

P(x) = C(n, x)pxqn-x

to compute the probabilities of all of the possible numbers of successes. The possible numbers of successes will be all of the whole numbers from 0 to 5, the number of trials.

Now that we have the probability distribution, we can find the expected value or the mean. To compute the expected value of the number of correcrt answers, multiply the number of correct answers by the probability ofgetting that many answers right, and add up the products.

So the expected number she got right would be 1/4 of 5 or 5/4. Now that we have the expected value or the mean, we can find the deviations. The variance is the expected value of the squares of the deviations.

So the standard deviation is the square root of 15/16 or approximately 0.9862.

The histogram looks like

which looks something like a bell shaped curve, except that left side of the graph is scrunched up against the end. This illustrates that bell shaped curve techniques do not do well when the mean is within two standard deviations from the ends of the horizontal axis.

For the probability that she got at least 2 right add up the probabilities of 2, 3, 4, and 5

270/1024 + 90/1024 + 15/1024 + 1/1024

= 376/1024

= 47/128

In this case it would be easier to compute the probability of the complementary event: none or one right.

243/1024 + 405/1024

= 648/1024

= 81/128

Our answer is the complementary probability which is computed by subtracting from 1, and it will check.

If we use the bell shaped curve techniques for these problems, the area under the bell shaped curve would be approximately 0.398. 47/128 = 0.367. Even though we are not expecting too close of agreement because the sample size is so small, and the mean is within two standard ddeviations of one of the endpoints, we do get not too bad of a ball park estimate.