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. :[
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
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.
X = 38 => Z = (38 - 50)/5 = -12/5 = -2.4
X = 62 => Z = (62 - 50)/5 = 2.4.
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 .....?