Question

You are to take a multiple-choice exam consisting of 64 questions with 5 possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let x represent the number of correct responses on the test. (a) What is your expected score on the exam? (Hint: Your expected score is the mean value of the x distribution.)(b) Compute the variance and standard deviation of x. Variance =Standard deviation =

Answer 1

Answer:

(a) The expected score is 12.8

(b) The standard deviation is 3.2 and variance is 10.24

Step-by-step explanation:

Consider the provided information.

You are to take a multiple-choice exam consisting of 64 questions with 5 possible responses to each question.

Here n=64 p=1/5 and q=1-1/5=4/5

Part (a)  we need to find the expected score on the exam.

Expected = np

Expected score = number of questions × P(right)

$Score = 64 \times \frac{1}{5} = 12.8$

Hence, the expected score is 12.8

Part (b)  Compute the variance and standard deviation of x.

Standard Deviation: $\sigma =\sqrt{npq}$

Now calculate the standard deviation as shown:

$\sigma =\sqrt{64\times \frac{1}{5}\times \frac{4}{5}}$

$\sigma =\sqrt{\frac{256}{25}}$

$\sigma =\frac{16}{5}=3.2$

Variance: $\sigma^2 =npq$

$\sigma^2 =64\times \frac{1}{5}\times \frac{4}{5}$

$\sigma^2 =\frac{256}{25}$

$\sigma^2 =10.24$

Hence, the standard deviation is 3.2 and variance is 10.24