# Another question on normal distribution

• Sep 14th 2010, 11:48 PM
mngeow
Another question on normal distribution
http://img815.imageshack.us/img815/632/hardassshit.jpg

I have no idea how to do these questions,I only know parts a(i) and b(i).
Also,are a(iii) and b(ii) conditional probability questions?
• Sep 15th 2010, 12:00 AM
Unknown008
Find the probability (percentage) of staff rated as poor.

Take 1 - P(poor)

This gives you the sum percentages of staff with fair and good ratings.

Then, divide this by 2 to get the percentage of fair and good ratings.

So, now depending on what type of z table you have, you can proceed with two different ways.
Find the z value for P(X > z) = P(Good)

or Find the z value for P(X < z) = P(Poor) + P(Fair)

When you got the z value, convert it into X.
• Sep 15th 2010, 12:08 AM
MathoMan
a(ii) Solving a(i) you know that the probability for a salesman to be rated poor equals 0.1288. Hence, 1-0.1288 of chance is left to be split among rates "fair" and "good". Meaning that there is an equal chance for a salesman to be rated "fair" and "good", each rate belonging the same probability (1-0.1288)/2=0.4356. So the surface under the probability density function is split into three parts: the first (leftmost) one measures 0.1288, and middle and the rightmost part each measuring 0.4356. All you have to do is find the value a having the property that 1-P(X<a)=0.4356. Can you do that?
• Sep 15th 2010, 12:11 AM
Unknown008
For the second one, you know that:

X~B(500, 0.98)

You get $\displaystyle Sd(x) = \sqrt{500(0.98)(0.02)}$

Hence,

X~N(np, npq)

X~N(490, 9.8)

This means, on average, 490 persons get the SMS within 3.5 seconds with a standard deviation of 9.8 people. If there are a total of 500 persons, more than 12 get the SMS after 3.5 seconds, that means there less than 488 people get the SMS within 3.5 s.

$\displaystyle P(X < 488) = P(Z < \frac{488 - 490}{\sqrt{9.5}})$

I think you can complete this now (Happy)
• Sep 15th 2010, 12:46 AM
mngeow
The value of Z is (1-P(poor))/2 right?
• Sep 15th 2010, 12:54 AM
Unknown008
hmm...

P(Poor) = 0.1288

P(Fair) + P(Good) = 1- 0.1288 = 0.8712

P(Good) = 0.8712/2 = 0.4356

P(X > z) = 0.4356

P(X < z) = 1-0.4356 = 0.5644

The z value corresponding to this probability is 0.162 from my z table.

Then, I use:

$\displaystyle z = \dfrac{x - \mu}{\sigma}$

$\displaystyle 0.162 = \dfrac{x-50}{5.3}$

$\displaystyle x = 50.85$

Hence, a staff needs at least a score of 50.85 to be classified as 'Good'.
• Sep 15th 2010, 01:11 AM
mngeow
ooh.I get it now.Didn't know it could be done that way.
Can I use X~N(500,0.02) and calculate by using P(X>12)?
• Sep 15th 2010, 01:15 AM
Unknown008
For the second one, yes, but be careful!

X ~ B(500, 0.02)

Thus,

X ~ N(10, 9.8)
• Sep 15th 2010, 01:15 AM
mngeow
Quote:

Originally Posted by Unknown008
hmm...

P(Poor) = 0.1288

P(Fair) + P(Good) = 1- 0.1288 = 0.8712

P(Good) = 0.8712/2 = 0.4356

P(X > z) = 0.4356

P(X < z) = 1-0.4356 = 0.5644

The z value corresponding to this probability is 0.162 from my z table.

Then, I use:

$\displaystyle z = \dfrac{x - \mu}{\sigma}$

$\displaystyle 0.162 = \dfrac{x-44}{5.3}$

$\displaystyle x = 44.9$

Hence, a staff needs at least a score of 44.9 to be classified as 'Good'.

I got that answer too,but its wrong.The answer key says its 50.8
• Sep 15th 2010, 01:17 AM
Unknown008
Oops, I typed in the wrong mean in my previous post.

$\displaystyle 0.162 = \frac{x - 50}{5.3}$

$\displaystyle x = 50.85$
• Sep 15th 2010, 01:54 AM
mngeow
Oh ok.One more thing I'm not sure about,if P(Good)=0.4356,why can't I find the x value of that?Why is P(X < z) = 1-0.4356 = 0.5644 required?
• Sep 15th 2010, 02:21 AM
Unknown008
My z table gives the probabilities of z values to the left of the z value. I cannot read the probability of the z value given the probability after that value.

That is, I cannot find the z value for P(X > z), but I can find for P(X < z).

However, I could take the value directly, but this will require that I also invert the sign of the z value later. To avoid confusion, I already take the precautions beforehand.

The third part is some sort of conditional probability.

$\displaystyle P(Good| >44) = \dfrac{P(Good)}{P(X > 44)}$
• Sep 15th 2010, 02:26 AM
mngeow
Oh,no wonder it was like that.Thanks a bunch!