# Math Help - Normal distribution problem

1. ## Normal distribution problem

Suppose that the distribution of marks on an exam is closely
described by a normal curve with a mean of 65. The 84th percentile of this distribution is 75.
(a) What is the 16th percentile?
(b) What is the approximate value of the standard deviation of exam
marks?
(c) What z score is associated with an exam mark of 50?
(d) What percentile corresponds to an exam mark of 85?
(e) Do you think there were many marks below 35? Explain.

These are some exercises the prof gave us. That's all of the information given.

Is that enough info to solve the problems? I'm having some difficulty getting started.

2. Originally Posted by BrownianMan
Suppose that the distribution of marks on an exam is closely

described by a normal curve with a mean of 65. The 84th percentile of this distribution is 75.
(a) What is the 16th percentile?
(b) What is the approximate value of the standard deviation of exam
marks?
(c) What z score is associated with an exam mark of 50?
(d) What percentile corresponds to an exam mark of 85?
(e) Do you think there were many marks below 35? Explain.

These are some exercises the prof gave us. That's all of the information given.

Is that enough info to solve the problems? I'm having some difficulty getting started.
You first need to calculate the standard deviation of the distribution:

Sinec the 84th percentile is 75 you know that Pr(X < 75) = 0.84. Now find the value of z* such that Pr(Z < z*) = 0.84 (where Z follows the standard normal distribution). Substitute that value into $Z = \frac{X - \mu}{\sigma}$: $z^* = \frac{75 - 65}{\sigma}$. Solve for $\sigma$.

Now you can attempt the other questions in the usual way (review your class notes and textbooks for examples).

3. So would 12.5 be the right answer for standard deviation?

4. Originally Posted by BrownianMan
So would 12.5 be the right answer for standard deviation?
No.

If you show all your work I will review it for mistakes.

5. The thing is, we haven't learned about probability distributions yet. We were only briefly introduced to the "normal curve" and the sample z-score calculated by (x - mean)/standard deviation. So I don't exactly understand your explanation, particularly this part: "Pr(Z < z*) = 0.84."

6. Originally Posted by BrownianMan
The thing is, we haven't learned about probability distributions yet. We were only briefly introduced to the "normal curve" and the sample z-score calculated by (x - mean)/standard deviation. So I don't exactly understand your explanation, particularly this part: "Pr(Z < z*) = 0.84."
Then perhaps you should revisit the question and this thread after you have learned a bit more. It is pointless trying to explain how to do things that you haven't learned yet.

7. Ok, I made another attempt. For z*, I looked at the z-score table and saw that the z-score that corresponds to P(Z < z*)=0.84 is 1.00. So then I substituted 1 into the equation z=(75-65)/st dev, and got

1 = 10/st dev, so therefore st dev is 10.

Is that right?

8. This is what I have so far:

a) 55%

b) 10

c) -1.5

d) 98th percentile

e) No, not many are below 35, because 99.7% fall between 35 and 95.

Can anyone let me know if this is correct, or if I have made any errors?

Thanks.

9. Originally Posted by BrownianMan
Ok, I made another attempt. For z*, I looked at the z-score table and saw that the z-score that corresponds to P(Z < z*)=0.84 is 1.00. So then I substituted 1 into the equation z=(75-65)/st dev, and got

1 = 10/st dev, so therefore st dev is 10.

Is that right?
Within the accuracy of your tables, that looks OK (I get z* = 0.9945 from my CAS).

10. Originally Posted by BrownianMan
This is what I have so far:

a) 55% Mr F says: This is 55, not 55%.

b) 10

c) -1.5

d) 98th percentile

e) No, not many are below 35, because 99.7% fall between 35 and 95.

Can anyone let me know if this is correct, or if I have made any errors?

Thanks.
It looks OK, apart from a minor correction to part (a).