Answer:
a) Expected score on the exam is 12.8.
b) Variance 10.24, Standard deviation 3.2
Step-by-step explanation:
For each question, there are only two possible outcomes. Either you guesses the answer correctly, or you does not. The probability of guessing the answer of a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The variance of the binomial distribution is:
![V(X) = np(1-p)](https://tex.z-dn.net/?f=V%28X%29%20%3D%20np%281-p%29)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
64 questions.
So ![n = 64](https://tex.z-dn.net/?f=n%20%3D%2064)
5 possible answers, one correctly, chosen at random:
So ![p = \frac{1}{5} = 0.2](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B1%7D%7B5%7D%20%3D%200.2)
(a) What is your expected score on the exam?
![E(X) = np = 64*0.2 = 12.8](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%20%3D%2064%2A0.2%20%3D%2012.8)
(b) Compute the variance and standard deviation of x. Variance =Standard deviation
![V(X) = np(1-p) = 64*0.2*0.8 = 10.24](https://tex.z-dn.net/?f=V%28X%29%20%3D%20np%281-p%29%20%3D%2064%2A0.2%2A0.8%20%3D%2010.24)
Variance 10.24
![\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{64*0.2*0.8} = 3.2](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B64%2A0.2%2A0.8%7D%20%3D%203.2)
Standard deviation 3.2