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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!

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Note that $1-(1-p)=p$ –  Daniel Littlewood Nov 11 '12 at 19:16

2 Answers 2

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.

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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.

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