1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Note that $1-(1-p)=p$ $\endgroup$ – preferred_anon Nov 11 '12 at 19:16
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Brilliant. Thanks a lot André! $\endgroup$ – DJDMorrison Nov 11 '12 at 16:17
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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