# Thread: Trying to solve a problem

1. ## Trying to solve a problem

The lenghts of pins produced bya machine follow anormal distribution with mean2.54cm and standard deviation 0.04cm.A pin is rejectedif its lenght is lest than 2.44cm or more than 2.60cm.
Find the percentage of pins that are accepted. I did this
upper limit =2.60-2.54/0.04 =1.5
lower limit=2.44-2.54/0.04 =-2.5 P(z)1.5-P(z)-2.5
=0.9332-(1-0.9938)
=92.7%
Is this right. The second part I need help with
If it is decidedthat 2.5%of the pins are to be rejected because they are to long and 2.5% because they are too short ,determine the new range of
acceptable lenghts. Could somebdy explain how to do this. Thank,s

2. First one is right.

In the second question, I donīt understand exactly what you mean.
But if you mean that the total percentage of pins rejected is 2.5%, then you need that 1.25% are too long, and 1.25% too short.

That is, $\displaystyle (1-x)=0.0125 \, \Rightarrow x = 0.9875$
From table we see that 0.9875 corresponds to 2.24, thus:

$\displaystyle \frac{t}{0.04} = 2.24 \, \Rightarrow t = 0.0896$

$\displaystyle \left[ 2.54-0.0896 \, , \, 2.54+0.0896 \right]$