Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could someone go over these calculations and tell me where I'm going wrong please. It's to do with normal distribution.

The question: In a factory, the packets of sweets produced are supposed to contain 1kg each. It has been found that the weights are normally distributed with mean 1.01kg and standard deviation 0.009kg. Find, to 1 d.p, the percentage of packets above the nominal 1kg weight.

So I need to find $P(Z>1)$

If I put it into the formula I get:

$$Z = \frac{1 - 1.01}{0.009}$$ $Z = - 1.1$ recurring

And I get a bit stuck here because to use my normal distribution chart, I need to make it so $P(Z < z)$, but its $P(Z > z)$ at the moment. Therefore I would just do $1 - P(Z < z)$. However, it's also a negative so I would have to do $1-P(Z < z)$ again.

Any help would be much appreciated!

share|cite|improve this question
Note that $1-(1-p)=p$ – Daniel Littlewood Nov 11 '12 at 19:16
up vote 1 down vote accepted

You want $\Pr(Z \gt -a)$, where $a$ is a positive constant. By symmetry of the standard normal, $$\Pr(Z\gt -a) =\Pr(Z\lt a).$$ This should be directly available from your tables for the cdf of the standard normal.

share|cite|improve this answer
Brilliant. Thanks a lot André! – DJDMorrison Nov 11 '12 at 16:17

p(x>1)=p[z>(x-u)/s] where x=1, u=1.01, and s=0.009 by substituting the values and simplifying, p(x>1)=p(z>-1.11) from the normal table, z-values are taken about 0 whether +tive or -tive they've the same value. Note that z-value for either +tive or -tive infinity is 0.5 Therefore, p(z>-1.11)=0.5+0.3665=0.8665 if you 're confused over this, consider finding the area between -1.11 to infinity on a number line where each of them are measured from 0 -1.11 to 0= a(e.g) 0 to +infinity = b -1.11 to +infinity = a+b where a and b are the values of -1.11 and +infinity respectively from the standard normal table.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.