1. ## Normal Distribution

Two questions any help would be great

1. There is a normal distributed class average of 70 and a standard deviation of 10. So what would be the 20,60,80 percentiles???

2.There is a normal distributed class with average(mean) of 71 and a standard deviation of 11.Find the quartiles(Q1,Q2,Q3)?

2. Originally Posted by someone21
Two questions any help would be great

1. There is a normal distributed class average of 70 and a standard deviation of 10. So what would be the 20,60,80 percentiles???

2.There is a normal distributed class with average(mean) of 71 and a standard deviation of 11.Find the quartiles(Q1,Q2,Q3)?
what have you tried?

Find z such that:

Pr(Z<=z) = 0.20
Pr(Z<=z) = 0.60 = 0.50+0.10 = 1.00 - 0.40
Pr(Z<=z) = 0.80 = 0.50+0.30 = 1.00 - 0.20

3. Originally Posted by TKHunny
what have you tried?

Find z such that:

Pr(Z<=z) = 0.20
Pr(Z<=z) = 0.60 = 0.50+0.10 = 1.00 - 0.40
Pr(Z<=z) = 0.80 = 0.50+0.30 = 1.00 - 0.20
SOrry but i just dont understand it how can be find the z score when we dont know what x value to plug in??

anyways thanks for trying

4. Originally Posted by someone21
SOrry but i just dont understand it how can be find the z score when we dont know what x value to plug in??

anyways thanks for trying
You have to use the concept of the inverse normal, which is what TKH was trying to push you towards.

For example:

$\Pr (Z \leq z) = 0.20 \Rightarrow \Pr( Z > -z) = 0.20 \Rightarrow \Pr(Z < -z) = 0.8 \Rightarrow -z = 0.842$

using the ubiquitous four-figure math tables. Therefore z = -0.842.

But you also know that $Z = \frac{X - \mu}{\sigma}$.

Therefore $0.842 = \frac{x - 70}{10} \Rightarrow x = ....$

5. Part 1a)

The 20th percentile in test scores is the test score which is surpassed by only 80% of the class. Likewise, the 60th percentile in test scores is the test score which is surpassed by only 40% of the class. Same deal with 80th percentile which is only surpassed by 20% of the class.

Now, let's call x the 20th percentile test score. How do we find x? Use the z-score equation: z = $\frac{x - \mu}{\sigma}$

We want the value of z that corresponds to .20. This value can be found using an online statistics applet (found in 30 seconds using google), or by reading your z-table (probably distributed countless number of times by your teacher and or in your text book). You will find that z = -0.84

So, plug and chug to get x: -0.84 = $\frac{x - 70}{10}$

Go from there and note that quartiles are the same question, just simply substitute for example, the first quartile with 25th percentile.