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The grades in a maths exam followed an aproximately normal distribution with mean value 112 and standard deviation 40 (in points. In this grading system the grades range from 0-200, 200 being the highest score).

In 10 students who took that exam, what is the probability that exactly 5 scored between 72 and 152 points?

a) 0.04

b) 0.12

c) 0.24

d) 0.34

Here is what I tried:

  • Using the normalcdf function in my calculator, I know that the probability of scoring between 72-152 is $\approx .682689$

Then I did this:

$$.682689^5 \approx .148$$

Which is aproximately b), which my book says is the correct answer.

But this doesn't look correct. Did I do something wrong? How do I solve this correctly? And also, how does knowing the total number of students influence the resolution of this problem?

2 Answers 2

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The problem says exactly 5 scored between 72 and 152. You need to take into account the other 5 people as well.

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    I tried $.682689^5\cdot (1-.682689)^5 \approx .0004$. What did I do wrong? – 2017-02-12
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    You need to take into account the number of ways the 10 people can answer that way (i.e. 10 choose 5 = 252). I get $252*.682689^5*(1−.682689)^5 = 0.1202095$. – 2017-02-12
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    Okay, I have one more question. What is $.682689^5*(1−.682689)^5$ the probability of? – 2017-02-12
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    You can think of the problem space in simple terms. I have ten balls and two buckets to put them in (balls being students and bucket one being scored between 72 and 152 and bucket two is they score outside that range). How many ways can I distribute the balls into the buckets so that exactly 5 are in both ( order doesn't matter so: n choose k = 10 choose 5 = 252). The probability of each of those 252 ways is the following: .682689^5∗(1−.682689)^5. You then need to sum all 252 cases together, but since they all have the some probability, we can just times it by 252. – 2017-02-12
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    Alright, so .682689^5∗(1−.682689)^5 is the probability of all 5 students of one group of those 252 having scored within the wanted range and .682689^5∗(1−.682689)^5*252 is the sum of that. I understand now, thanks. – 2017-02-12
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That's the probability of 1 single person scoring into that range, you gotta do it for exactly 5.