Consider a multiple-choice examination with 50 questions....

Consider a multiple-choice examination with 50 questions. Each question has four possible answers. Assume that a student who has done the homework and attended lectures has a $75 \%$ probability of answering any question correctly. a. A student must answer 43 or more questions correctly to obtain a grade of A. What percentage of the students who have done their homework and attended lectures will obtain a grade of $\mathrm{A}$ on this multiple-choice examination? b. A student who answers 35 to 39 questions correctly will receive a grade of $C .$ What percentage of students who have done their homework and attended lectures will obtain a grade of $C$ on this multiple-choice examination? c. A student must answer 30 or more questions correctly to pass the examination. What percentage of the students who have done their homework and attended lectures will pass the examination? d. Assume that a student has not attended class and has not done the homework for the course. Furthermore, assume that the student will simply guess at the answer to each question. What is the probability that this student will answer 30 or more questions correctly and pass the examination?

Consider a multiple-choice examination with 50 questions. Each question has four possible answers. Assume that a student who has done the homework and attended lectures has a $75 \%$ probability of answering any question correctly.
a. A student must answer 43 or more questions correctly to obtain a grade of A. What percentage of the students who have done their homework and attended lectures will obtain a grade of $\mathrm{A}$ on this multiple-choice examination?
b. A student who answers 35 to 39 questions correctly will receive a grade of $C .$ What percentage of students who have done their homework and attended lectures will obtain a grade of $C$ on this multiple-choice examination?
c. A student must answer 30 or more questions correctly to pass the examination. What percentage of the students who have done their homework and attended lectures will pass the examination?
d. Assume that a student has not attended class and has not done the homework for the course. Furthermore, assume that the student will simply guess at the answer to each question. What is the probability that this student will answer 30 or more questions correctly and pass the examination?

Key Concepts

Normal Approximation

When the number of trials is sufficiently large, the binomial distribution can be approximated by the normal distribution. This approximation simplifies the computation of cumulative probabilities, especially when determining probabilities over intervals or thresholds, such as passing criteria or grade cutoffs.

Cumulative Distribution Function (CDF)

The cumulative distribution function for a binomial random variable aggregates probabilities up to a certain number of successes. It is used to determine the probability of achieving at least or at most a given score by summing the probabilities of individual counts, which is essential for assessing grade thresholds.

Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent trials where each trial has the same probability of success. It is fundamental for answering questions involving counts of correct answers, as each question can be modeled as a trial with a success (correct answer) or failure (incorrect answer).

Independent Bernoulli Trials

Each individual trial (or question) in the exam has only two outcomes (correct or incorrect) with a given probability of success, and the result of any one trial does not affect the others. This independence is a key assumption that underpins the use of the binomial distribution in these types of probability problems.

Probability Mass Function (PMF)

The probability mass function of a binomial random variable gives the probability of obtaining exactly a certain number of successes in a fixed number of trials. It is used to calculate the chance of answering exactly k questions correctly and is a building block for further probability calculations, such as cumulative probabilities.

Transcript

00:01 For this problem, we are to consider a multiple choice examination with 50 questions that represents our n and is equal to 50.

00:09 Now, each question carries four possible answers.

00:13 And the fact that a student who attends lectures and does the homework has a 75 % chance of getting an answer correctly, that is p, which is equal to 0 .75.

00:25 For the first question, a student is graded a when he answers 43 questions or more.

00:34 We have to find a percentage of students who, do their homework and attend classes and will obtain grade a in this particular examination.

00:44 So to get started, we need to identify the kind of distribution the scenario follows.

00:51 So the scenario follows a binomial distribution.

01:01 This is because there are two possible outcomes in an examination, that is a pass or a fail.

01:08 The probability of passing is 0 .75.

01:11 The probability of failing would definitely be 1 minus p, that is 0 .25.

01:15 And we also have number of trials that is 50 and the probability in each trial is constant.

01:26 Now in computing the probabilities, n is large and therefore the normal approximation to binomial will do a very good job.

01:38 So to confirm the fact that we can use the normal approximation to binomial, let's compute n times p.

01:46 And that is equal to 50 times 0 .75 and that is equal to 37 .5.

01:54 Now since n times p is greater than 5 then this scenario can be approximated using the normal distribution.

02:08 So for the first problem you have to find a percentage of students that can answer 43 questions or more.

02:15 That is the probability that x is greater than or equal to 43.

02:21 Now in approximating a discrete probability such as a binomial distribution using a continuous probability distribution like the normal distribution we use what we call continuity and applying continuity this becomes the probability that x is greater than or equal to 43 minus 0 .5 which is equal to 42 .5.

02:58 Now we need to compute the z score for this value and the formula is given as z is equal to x minus the mean divided by standard deviation.

03:25 So we need to identify the mean and standard deviation in this situation.

03:33 Now the mean or m is equal to n times p that is 50 times 0 .75 and it gives us 37 .5.

03:49 The standard deviation however is equal to the square root of n times p multiplied by 1 minus p now the square root of now n times p is already 37 .5 and then 1 minus p that is 1 minus 0 .75 that is 0 .25 now the standard deviation becomes 3 .062 so we go ahead and compute the z score so you have the probability that x becomes z is great to than or equal to the x value that is 42 .5 minus the mean 37 .5 divided by the standard deviation 3 .062.

04:53 And i decide the probability that z is greater than or equal to 1 .6.

05:05 To compute this probability is equal to 1 minus a normal value, 1 minus a normal value of 1 .6 .3.

05:30 That is equal to 1 minus then we find the probability of the percentage that corresponds to 1 .63 from the standard z table that is 0 .9484 and this gives us 0 .0 .056 and move on to the next question now when a student answers 35 to 39 questions he is rewarded the grade of c let's find a percentage of students who have done their homework and attended lectures and will obtain a grade of c on this particular examination.

06:26 So you can write that as the probability that 35 is less than or equal to x.

06:33 It is less than or equal to 39.

06:38 Now applying continuity, we have 35 minus 0 .5...

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