An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,

  1. Compute the expected mean for the student score
  2. Compute the standard deviation for the student score
  3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
  4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam :  What is the expected success rate?  How do you expect the proportion of students who will score less or equal to 20?

If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?

a. What is the expectation of this student's degree?

b. what is the standard deviation of this student's grade?

b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?

  1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
  • 4
    $\begingroup$ Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. $\endgroup$ – saulspatz May 21 at 0:59
  • $\begingroup$ thanks I did it $\endgroup$ – Nidal May 21 at 1:29

It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.

For a binomial random variable $X$ with $n = 50$ and $p = 1/5,$ you can use a normal approximation to binomial to get reasonable approximate answers to these questions. (You will get somewhat better approximations, if you use a continuity correction, but even then you can't expect to get more than two or three places of accuracy.)

Alternatively, you can use statistical software (or perhaps a statistical calculator) to to get answers exact to many decimal places.

Below are some answers from R statistical software, in which dbinom and pbinom, designate a binomial PDF and CDF, respectively. I assume each correct guess counts 2 points and that there is no penalty for incorrect guesses.

$P(X \ge 30) = 1 - P(X \le 29) \approx 0$

1 - pbinom(29, 50, .2) 
[1] 6.936747e-10

$P(X \le 10) = .5836.$

pbinom(10, 50, .2)
[1] 0.5835594

$P(X = 0) = .8^{50}.$

dbinom(0, 50, .2);  .8^50.
[1] 1.427248e-05
[1] 1.427248e-05

A student who has studied half of the material should have probability $p = 0.6$ of answering each question correctly. Why? Let $Y$ be the number of correctly answered questions.

$P(Y \ge 30) = 1 - P(Y \le 29) = 0.5610.$

 1 - pbinom(29, 50, .6)
 [1] 0.5610349

Using a normal approximation. The number $Y$ correct has $E(Y) = np = 50(.6) = 30,$ $Var(Y) = 12,$ $SD(Y) = 3.4641.$

Thus $Y \stackrel{aprx}{\sim} \mathsf{Norm}(\mu=30, \sigma=3.4641)$ and (standardizing) $Z = \frac{Y - \mu}{\sigma} \stackrel{aprx}{\sim} \mathsf{Norm}(0, 1).$

Hence, $P(Y \ge 30) = 1 - P(Y < 29.5) \approx 0.557.$ (If you standardize and use printed normal tables you will get about the same answer; not quite because of the rounding involved in using the table.)

 1 - pnorm(29.5, 30, 3.4641)
 [1] 0.5573831

The figure below shows the PDF of $\mathsf{Binom}(50,.6)$ (bars) along with the density function of $\mathsf{Norm}(30, 3.4641)$ (blue curve). The probability $P(Y \ge 30)$ is the sum of the heights of the lines to the right of the dotted red line. (This probability is approximated by the area under the normal density curve to the right of this vertical line.)

enter image description here

The R code for the figure is shown below:

 y = 0:50;  PDF = dbinom(y, 50, .6)
 plot(y, PDF, type="h", lwd=2, main="PDF of BINOM(50,.35) with Normal Approx.")
   curve(dnorm(x, 30, 3.4641), 0, 50, add=T, col="blue")
   abline(h=0, col="green2");  abline(v=0, col="green2")
  • $\begingroup$ that was helpful,,, when he studied half of the material why p=0.6 I didn't get it. $\endgroup$ – Nidal May 21 at 10:55
  • $\begingroup$ when he has no prior knowledge p was 1/5 (like one option will be right from the five options), now when he has studied half of the material why p is 3/5=0.6. $\endgroup$ – Nidal May 21 at 14:06
  • $\begingroup$ S = studied, N = not. P(S)=P(N) = 1/2. P(Corr) = 1P(S) + (1/5)P(N) = .5 + .5(.2) = .5+.1 = .6. Law of total probability. $\endgroup$ – BruceET May 21 at 15:54
  • $\begingroup$ Unsolicited, so likely unwelcome advice: I have had a look at your history of activity on this site. In my opinion, although you ask some nice questions, you reveal much less of your work and thinking than optimal. (Even those bits often only on demand.) Looks as if you are taking a fine and well-taught course. Outsourcing too much of the problem solving is not a path toward effective learning. $\endgroup$ – BruceET May 21 at 16:20
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    $\begingroup$ We can't be sure there are exactly 25 questions from the parts the student studies. The way I read it, there are on average 25 such questions. // Your way and mine both give expected nr correct as 30, but for slightly different distributions. Mine BINOM(50, .6), yours 25 + BINOM(25, .2). Same mean, mine has larger variance. $\endgroup$ – BruceET May 21 at 19:33

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