# Help with normal Distribution

• Oct 21st 2011, 12:57 AM
sparkeeuk
Help with normal Distribution
Hi, I wondered if someone on here could kindly help with one of my maths questions for my course as I can't seem to get my head round this and am hopin someone could put me in the right direction if possible, heres the question.......

The Lengths of pins produced by a machine follow a normal distribution with mean of 2.54cm and a standard deviation of 0.04cm. A pin is rejected if the length is less than 2.44cm or more than 2.60cm

a) What is the percentage of pins accepted

b) If it is decided that 2.5% of the pins are to be rejected because they are too short and 2.5% because they are too long, what is the new range of acceptable lengths

Would really appreciate any help on this (Worried)
• Oct 21st 2011, 03:20 AM
mr fantastic
Re: Help with normal Distribution
Quote:

Originally Posted by sparkeeuk
Hi, I wondered if someone on here could kindly help with one of my maths questions for my course as I can't seem to get my head round this and am hopin someone could put me in the right direction if possible, heres the question.......

The Lengths of pins produced by a machine follow a normal distribution with mean of 2.54cm and a standard deviation of 0.04cm. A pin is rejected if the length is less than 2.44cm or more than 2.60cm

a) What is the percentage of pins accepted

b) If it is decided that 2.5% of the pins are to be rejected because they are too short and 2.5% because they are too long, what is the new range of acceptable lengths

Would really appreciate any help on this (Worried)

a) Calculate Pr(2.44 < X < 2.60) = Pr(X < 2.60) - Pr(X < 2.44). How these are calculated will depend on how you've been taught. Your textbook and class notes will have many examples on how to calculate these two probabilities. Remember to convert the answer into a percentage.

b) Find the value of a such that Pr(X < a) = 0.025, find the value of b such that Pr(X > b) = 0.025. You need to do an inverse normal probability calculation. Again, your textbook and class notes will have many examples to follow.
• Oct 21st 2011, 04:47 AM
salim
Re: Help with normal Distribution
Hi,
The problem is about a random variable X that has a normal distribution, such that:

A : the event that represent the accepted pins $\displaystyle \left ( 2.44\leqslant x\leq 2.66 \right )$
B : the event that represent the rejected pins $\displaystyle \left ( x< 2.44\right )and\left ( x> 2.66 \right )$
It's clear that B = A' then P(B) = 1 - P(A).
and :
$\displaystyle P\left ( A \right )= \int_{2.44}^{2.6}\frac{1}{\sqrt{2\pi \left ( 0.04 \right )^{2}}}E^{-\frac{\left ( x - 2.54 \right )^{2}}{2\left ( 0.04 \right )^{2}}}dx$
P(A) = 0.99244 => %A = 99.244% accepted
• Oct 21st 2011, 01:00 PM
mr fantastic
Re: Help with normal Distribution
Quote:

Originally Posted by salim
Hi,
The problem is about a random variable X that has a normal distribution, such that:

A : the event that represent the accepted pins $\displaystyle \left ( 2.44\leqslant x\leq 2.66 \right )$
B : the event that represent the rejected pins $\displaystyle \left ( x< 2.44\right )and\left ( x> 2.66 \right )$
It's clear that B = A' then P(B) = 1 - P(A).
and :
$\displaystyle P\left ( A \right )= \int_{2.44}^{2.6}\frac{1}{\sqrt{2\pi \left ( 0.04 \right )^{2}}}E^{-\frac{\left ( x - 2.54 \right )^{2}}{2\left ( 0.04 \right )^{2}}}dx$
P(A) = 0.99244 => %A = 99.244% accepted

It is highly unlikely that the OP will be calculating probabilities from the pdf.

The OP will be using tables, s/he needs to get help from his/her instructor on how to use those tables.