I can't seem to get the right answer for this problem...

Given that the population of scores is normally distributed with (mean=110) and (standard deviation=8), determine the following:

What is the percentage of scores that are between a score of 101 and 122?

2. Originally Posted by kelly3223
I can't seem to get the right answer for this problem...

Given that the population of scores is normally distributed with (mean=110) and (standard deviation=8), determine the following:

What is the percentage of scores that are between a score of 101 and 122?

If $X = 101$ then $Z = \frac{X - \mu}{\sigma} = \frac{101 - 110}{8} = z_1$.

If $X = 122$ then $Z = \frac{X - \mu}{\sigma} = \frac{122 - 110}{8} = z_2$.

(Your job by the way is to get the z-values ...)

Then $\Pr(101 < X < 122) = \Pr(z_1 < Z < z_2) = \Pr(Z < z_1) - \Pr(Z < z_2) = p$.

(Your job by the way is to get the value of p by calculating $\Pr(Z < z_1)$ and $\Pr(Z < z_2)$. I assume you've been taught how to do this using tables etc. ....)

The percentage of scores is therefore 100p.

If you need more help please show your working and say where you get stuck.

3. I'm still confused. I did the equation to find z1 and z2 and got -1.125 for z1 and 1.5 fo z2, but what do I do next to get the answer?

4. Originally Posted by kelly3223
I'm still confused. I did the equation to find z1 and z2 and got -1.125 for z1 and 1.5 fo z2, but what do I do next to get the answer?

Pr(z < -1.125) = Pr(z > 1.125) = 1 - Pr(z < 1.125).

Now calculate Pr(z < 1.125) (and hence Pr(z < -1.125)) and Pr(z < 1.5) and go back and read my first post for what to do.

5. ## I think I got it?

So, 1.13 = .3708 (according to our table)
1.50= .4332 ('' ")

Percentage between 101 and 122: (0.3708+0.4332)x100=80.4%

Is this correct?!