# 75th percentile of the normal distribution

1. Jun 9, 2017

### zak100

1. The problem statement, all variables and given/known data
The random variable X is normally distributed. The values 650 and 850 are at the 60th and 90th percentile of the distribution of X respectively. Find the value of the 75th percentile of the distribution of X.

2. Relevant equations

I found the above formula from internet.
where sigma is standard deviation, mu is mean

3. The attempt at a solution
I dont know the values of mu, sigma and Z. Somebody please guide me.

Zulfi.

2. Jun 9, 2017

### Ray Vickson

Use a table of standard normal distribution values to get the two values of $z$ for the 60th and 90th percentiles.

3. Jun 9, 2017

### .Scott

From a SD table look-up:
The 60th percentile puts 650 at a standard deviation of 0.2535.
So (650-u)/s = 0.2535.
Now do the same for 850 and you'll have two linear equations and two unknowns.

4. Jun 9, 2017

### zak100

Hi,
Thanks for your replies. I don't think that in the exam, I would be having access to this SD table. I have to do with out the table.
Is 75th percentile not between 90th and 60th percentile. (850 + 650)/2 = 750. Why is this wrong? I have seen an example which finds 15th percentile using the values of 10th and 20th percentile. Some body please guide me.

Zulfi.

5. Jun 9, 2017

### Ray Vickson

The 75th percentile of the normal distribution is NOT the average of the 60th and 90th percentiles. For the standard normal, the average of the 60th and 90th percentiles is $z_m = 0.7674493346,$ and this is the 77.86th percentile. The relation between the $z$-value and the probability value is not linear (that is, does not have a straight-line graph), so you cannot just take simple averages like you did.

Also: whoever told you that the 15th percentile is the average of the 10th and 20th percentiles is wrong; that gives you the 14.42th percentile.

If you do not have access to normal tables in an exam, all you can do is express the answer symbolically: if you let $a =$60th percentile and $b =$ 90th percentile of the standard normal, you can get $\mu$ and $\sigma$ in terms of $a, b$ and the given input numbers. Then if you let $c =$ 75th percentile of the standard normal, you can write a formula for the 75th percentile of $X$ in terms of $a, b$ and $c$.

6. Jun 13, 2017

### zak100

Hi,
I am attaching the diagram.

Can somebody please point out where 750 would lie in the attached graph and where the value of 75th percentile would lie? This would help me to find out whether 750 is greater or 75th percentile is greater.

Zulfi.

7. Jun 13, 2017

### Ray Vickson

No, there is technically no easy way for anybody to point out on the graph where the 75th percentile would be. Anyway, locating the 75th percentile is YOUR job, and you should do everything you can to finish the task. You have been told several times how to find the 75th percentile: consult a table of the standard normal distribution. You say that in an exam you would not have access to a "normal" table, but that is irrelevant here: you are posting a message, not writing an exam.

That being said, you can answer the question for yourself as to whether the 75thpercentile is less than or greater than 750. On the graph, the point at 750 cuts the area under the graph (between 650 and 850) into two parts. Just looking at the picture, do you think that the area between 650 and 750 is equal to the area between 750 and 850? Basically, the 75th percentile $p$ is determined by making the area between 650 and $p$ equal to the area between $p$ and 850.

On an exam, that type of analysis might be sufficient, even without having a normal table available.

8. Jun 18, 2017

### Staff: Mentor

You do not get asked such a question without a table (at least with 60, 75 and 90 as entries).
While it is possible to find approximate values for the quantiles from scratch (someone made the tables, by hand before computers were around) that is way too much work for an exam.