Thread: I need help with normal distributions...

Hello! :]

I've had no trouble with normal distribution questions until I got to this one...

Q. The mean of normal distribution is 50 and the standard deviation is 5. What percentage of the data:

a. lies between 38 and 62

(b+c+d are pretty much the same question, but with different numbers.)

e. Is less than 38.

I haven't worked with numbers that haven't fallen into any of the standard deviations. Could I have a little help please? I have no idea what to do. :[

Originally Posted by Hikaru

Hello! :]

I've had no trouble with normal distribution questions until I got to this one...

Q. The mean of normal distribution is 50 and the standard deviation is 5. What percentage of the data:

a. lies between 38 and 62

(b+c+d are pretty much the same question, but with different numbers.)

e. Is less than 38.

I haven't worked with numbers that haven't fallen into any of the standard deviations. Could I have a little help please? I have no idea what to do. :[

Have you been taught how to get the z-value? And have you been taught how to use tables of values of the standard normal distribution?

I have, but I've completely forgotten about it... I've been in and out of hospital recently, so I might have missed some vital notes here and there.

I guess I'll ask my teacher tomorrow on how to use Z-values, thanks =]

(But if it follows an easy formula, could you please post it anyway?)

Uh... I don't think we've used any tables... O_o

Originally Posted by mr fantastic

Have you been taught how to get the z-value? And have you been taught how to use tables of values of the standard normal distribution?

Hello,

The principle of using a z-table is as following :

If a random variable X is normally distributed, with mean m and standard deviation s, then the variable Z=(X-m)/s will follow a standard normal distribution, which representation will be the red curve here :


Note that P(X<x)=P(Z<(x-m)/s).

And there exists a table (http://www.science.mcmaster.ca/psych...e/z-table2.jpg) that gives you P(Z<z).

After that, it's all juggling with these values. See the table so that you may be able to understand a little more efficiently.

Originally Posted by Hikaru

I have, but I've completely forgotten about it... I've been in and out of hospital recently, so I might have missed some vital notes here and there.

I guess I'll ask my teacher tomorrow on how to use Z-values, thanks =]

(But if it follows an easy formula, could you please post it anyway?)

Uh... I don't think we've used any tables... O_o

Z = (X - mean)/sd.

X = 38 => Z = (38 - 50)/5 = -12/5 = -2.4

X = 62 => Z = (62 - 50)/5 = 2.4.

Therefore:

Pr(38 < X < 62) = Pr(-2.4 < Z < 2.4) = Pr(Z < 2.4) - Pr(Z < -2.4)

= Pr(Z < 2.4) - Pr(Z > 2.4)

using symmetry of the normal distribution

= Pr(Z < 2.4) - (1 - Pr(Z < 2.4))

= 2 Pr(Z < 2.4) - 1.

And you get Pr(Z < 2.4) using the standard normal distribution tables (unless you use technology).

Looking at your previous posted questions (and the replies), I'm surprised that you're having trouble here .....?