Binomial distribution and normal approximation -is my working correct ?
Q:A multiple-choice examination consists of 20 questions, for each of which the candidate is required to tick as correct one of three possible answers. Exactly one answer to each question is correct. A correct answer gets 1 mark and a wrong answer gets 0 marks. Consider a candidate who has complete ignorance about every question and therefore ticks at random. What is the probability that he gets a particular answer correct? Calculate the mean and variance of the number of questions he answers correctly.
The examiners wish to ensure that no more than 1% of completely ignorant candidates pass the examination. Use the normal approximation to the binomial, working throughout to 3 decimal places, to establish the pass mark that meets this requirement.
I'm only concerned about the last part of the question
The examiners wish to ensure that no more than 1% of completely ignorant candidates pass the examination. Use the normal approximation to the binomial, working throughout to 3 decimal places, to establish the pass mark that meets this requirement.
I'm only concerned about the last part of the question
Original post by JanaALEVEL
Q:A multiple-choice examination consists of 20 questions, for each of which the candidate is required to tick as correct one of three possible answers. Exactly one answer to each question is correct. A correct answer gets 1 mark and a wrong answer gets 0 marks. Consider a candidate who has complete ignorance about every question and therefore ticks at random. What is the probability that he gets a particular answer correct? Calculate the mean and variance of the number of questions he answers correctly.
The examiners wish to ensure that no more than 1% of completely ignorant candidates pass the examination. Use the normal approximation to the binomial, working throughout to 3 decimal places, to establish the pass mark that meets this requirement.
I'm only concerned about the last part of the question
The examiners wish to ensure that no more than 1% of completely ignorant candidates pass the examination. Use the normal approximation to the binomial, working throughout to 3 decimal places, to establish the pass mark that meets this requirement.
I'm only concerned about the last part of the question
I believe you are on the right lines. You have determined that if the pass mark was set at 12.1, then the probability of a student passing by random choice of answers would be 0.01 as required. However, the pass mark must be a whole number of questions, so we must round up to a pass mark of 13. (Rounding down to 12 would result in the probability of a "random" pass being above 0.01).
In an exam, I would suggest showing a bit more working, particularly a statement of your normal approximation X~N(20/3, 40/9). Also you should finish with a statement in context.
You can always check P(X > 11.5) and P(X > 12.5) using your calculator's cumulative normal distribution function.
(edited 4 years ago)
Where you wrote "mean mark" did you mean "pass mark"?
Notice the question says to work to 3 d.p.
Notice the question says to work to 3 d.p.
Original post by old_engineer
You can always check P(X > 11.5) and P(X > 12.5) using your calculator's cumulative normal distribution function.
You can always check P(X > 11.5) and P(X > 12.5) using your calculator's cumulative normal distribution function.
o I have a question, what would I do with these ?
Original post by MarkFromWales
Where you wrote "mean mark" did you mean "pass mark"?
Notice the question says to work to 3 d.p.
Notice the question says to work to 3 d.p.
ah sorry, yeah I meant pass mark.
for the 3 dp thing, if each question is worth 1 mark shouldn't the answer be in that context ? Or do you mean something else ?
Original post by JanaALEVEL
o I have a question, what would I do with these ?
If you have a Casio fx-991EX calculator, you can calculate P(X > 11.5) for the normal distribution by entering Menu 7 (Distribution) / Normal CD / then set Lower = 11.5, Upper = (say) 10,000, standard deviation = sqrt(40/9) and mean = 20/3. The answer is 0.0109, confirming that a pass mark of 12 is too low. Other scientific calculators will have similar functions.
(Original post by JanaALEVEL)ah sorry, yeah I meant pass mark.
It's just a detail but the question said "working throughout to 3 decimal places". I think that would apply to the 2.33 and the 12.1 in your answer, but not to the final answer which presumably has to be a whole number.
for the 3 dp thing, if each question is worth 1 mark shouldn't the answer be in that context ? Or do you mean something else ?
It's just a detail but the question said "working throughout to 3 decimal places". I think that would apply to the 2.33 and the 12.1 in your answer, but not to the final answer which presumably has to be a whole number.
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