1. normal distribution

This seems to be straight forward but I can't seem to work it out:

The error made when a certain instrument is used to measure the body length of a butterfly of a particular species is known to be normally distributed with mean 0 and standard deviation 1mm. Calculate to 3d.p, the probability that the error made when the instrument is used once is numerically less than 0.4mm.

The answer given in my textbook is 0.311

Can someone explain why?

Thanks.

2. Originally Posted by jedoob
This seems to be straight forward but I can't seem to work it out:

The error made when a certain instrument is used to measure the body length of a butterfly of a particular species is known to be normally distributed with mean 0 and standard deviation 1mm. Calculate to 3d.p, the probability that the error made when the instrument is used once is numerically less than 0.4mm.

The answer given in my textbook is 0.311

Can someone explain why?

Thanks.
This problem confuses me. Let me explain, because:
$\displaystyle \mbox{error }=|\mbox{actual value }-\mbox{ approximate value}|$.
From here we see that error is an absolute value thus it is always non-negative.
Thus, when you say that error be no less than .4 do you mean that the error be less than .4
thus, $\displaystyle |x|<.4$ which means $\displaystyle -.4<x<.4$ henceforth the probability you are trying to find is,
$\displaystyle P(-.4<x<.4)$
Or do you simply mean to say that error in this case the the difference between the actual and approximate values thus you can have a negative value. Meaning you are trying to find the probability $\displaystyle P(x<.4)$.

In either case let me do both problems.
Let us find the z-score from $\displaystyle 0<x<.4$ which is, $\displaystyle z=.4$. Looking up at the tables we have that the probability is, $\displaystyle P(0<x<.4)=.1554$.
Now if you say the first case, then we also need
$\displaystyle P(-.4<x<0)=.1554$ which is the same because we have symmetry arround the mean. In total we have $\displaystyle P(-.4<x<.4)=.1554+.1554=.3108\approx .311$
If it were the other case you need to add $\displaystyle P(x<0)=.5$ because the left hand side of the curve takes exactly half the values. In total you would then have .655