# How do you use the mode to find x% of y?

1. Dec 4, 2013

### semidevil

1. The problem statement, all variables and given/known data

Test scores are normally distributed. The mode of the score is 55%. If 75% of the students scored 65, what % of student scored 85?

2. Relevant equations and attempt at solution

Scores being normally distributed, I know that as the 'bell curve (that means the curve is distributed with mean 0, variance 1).'

I also know 'mode' is the peak of the bell curve, since mode is the most people who scored a 55.

I really don't know where and how to proceed from there.

25% of the students scored 65, I can do (x - 65)/ standard deviation = .25% (or phi .68). I get x = 65.68 students.

So around 66 of students scored 65...right?

What's the next step? I don't know how to use the mode in this problem.

2. Dec 4, 2013

### LCKurtz

I don't understand your statement of the problem. If 75% of the students scored 65, the mode couldn't be 55 because there are only 25% of the students left.

3. Dec 5, 2013

### Ray Vickson

Your understanding of the "bell curve" is wrong: it just means that the distribution is normal. The mean need not be 0 and the variance need not be 1.

4. Dec 5, 2013

### Staff: Mentor

In addition to what Ray and LCKurtz have already said, I suspect that the problem statement is incomplete or wrong.
If 75% of the students scored 65, how can the mode possibly be 55%? The mode is the score that occurs most often.

Is the 3rd sentence in what I quoted the exact statement or are some words missing? It would make more sense if it said something like this: "If 75% of the students scored 65 or below, ..."

5. Dec 5, 2013

### semidevil

yes, thank you. your right. I'm trying to remember a problem that I encountered and I thought those were the right numbers, but apparently, they are not.

so with that in mind, I guess what I'm more after is what the mode has to do with the problem and how it can be used to find the probability that x% of people scored y. I know that mode is the one that occurs the most, but besides that, I don't know what to do with it.

6. Dec 5, 2013

### Staff: Mentor

In a normal distribution, the mode equals the mean, so here the mean is 55. (Let's drop the % business for the scores for the sake of simplicity.)

If 75% of the students scored 65 or below ("or below" is my assumption as to what the problem is saying), then this says that 25% of the distribution is between 55 and 65. This gives you some idea of what the standard deviation is, which as Ray already said, doesn't have to be 1.

In the standard normal distibution, the mean is 0 and the standard deviation is 1. In a normal distribution, the mean and standard distribution can be whatever.

Your textbook might have approximate ranges for how much of a normal distribution is within 1 standard deviation of the mean, two s.d. of the mean, and so on.

7. Dec 5, 2013

### semidevil

thanks! that is exactly what I needed. I did not know that the mode equals the mean in this situation!

8. Dec 6, 2013

### LCKurtz

There is a difference between continuous densities and discrete probability functions. For continuous densities, you don't talk about x% of the people scoring y. The probability that anyone scores a particular value is 0. For the discrete case, the mode is indeed the outcome occurring most frequently. For a continuous density the mode is the value $x$ where the maximum of the density function is attained. The probability of the outcome $x$ is zero, unlike the discrete case. In the continuous case you are interested in the probability of the outcome being in some interval, not some particular value.