# Homework Help: Statistics: normal distribution

1. Jun 27, 2011

### k77i

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

Suppose that X is normally distributed with mean 95 and standard deviation 17.

A. What is the probability that is greater than 126.79?

B. What value of X does only the top 16% exceed?

2. Relevant equations

z= (X-(mean of x))/standard deviation

3. The attempt at a solution

using the equation, I found z= (126.79-95)/17=1.87

then using a statistical table, the value for P(z greater than 1.87) = 1 - P(z=1.87) = 0.0307

But this is wrong..

For B)

I considered that 84% must be the value which the top 16% must exceed.. then 84% of 95 is 88.42

Also wrong (well I wasn't really too sure about part B)

Any help would be appreciated!

2. Jun 27, 2011

### Dickfore

The tabular value for above that z-score seems correct. Your z-score seems correct to the given number of decimal places. How do you know that the answer under a) is "wrong"?

For B), you should find such a $Z_{0}$ that:

$$P(Z > Z_{0}) = 0.16 \Rightarrow P(Z < Z_{0}) = 0.84$$

Using that value for $Z_{0}$, solve for X in the expression:
$$Z = \frac{X - \mu}{\sigma}$$
where $\mu$ is the mean and $\sigma$ is the standard deviation.

3. Jun 27, 2011

### HallsofIvy

No, that is correct. Why do you say it is wrong?

95 is the mean- 50% are larger than 95!

Looking at a table of the normal distribution, N(x< a)= 0.84 for a= 1.0 (approximately).

$$\frac{z- 95}{17}= 1.0$$
solve for z.

4. Jun 27, 2011

### k77i

Thank you both for the help with part B. As for part A, since its a webwork assignment, it tells me when I enter an incorrect answer. I guess it's possible that there might be some sort of problem with the webwork.

5. Jun 28, 2011

### Dickfore

maybe you need to use the correct number of significant figures.

6. Jun 28, 2011

### k77i

I put all the numbers (0.0307).. The chart I have for the z-score only goes upto 4 decimal places so I'm not sure how much it's actually supposed to go up to

7. Jun 29, 2011

### Dickfore

Well, assuming the mean and the standard deviation are known to the same number of significant figures as 126.79, then the z-score is:
$$\frac{126.79 - 95}{17} = \frac{31.79}{17} = 1.870$$
to 4 significant figures (actually the quotient is exact). For this, the probability in question is:

1 - 0.9693 = 0.0307

So, it can't be that.

Are you sure this is the standard deviation and not the variance?

8. Jun 29, 2011

### k77i

Yes it is the standard deviation. I copied the question word for word..
In any case it's too late, the homework set is closed and apparently the accepted answer was 0307419079428789. I think I was supposed to use excel to find the exact value to the correct amount of significant figures..

9. Jun 30, 2011

lol.