Normal distribution of rods

  • Thread starter brad sue
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  • #1
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Hi, I have 2 problems I would like some help. It is about normal distribution(probability)

PROBLEM 1: Extruded plastic rods are automatically cut into lenghts of 6 inches. Actual lengths are normally distributed about a mean of 6 inches and their standard deviation is 0.06 inch.

1- what proportion of the rods have lenghts that are outside the tolerance limits of 5.9 and 6.1 inches?


Here I did:
p=F((6.1-6)/0.06)- F((5.9-6)/0.06)= F(1.67)-F(-1.67)=0.9525-0.0475=0.905

P(outside tolerance)=1-0.905=0.095

2- To what value does the standard deviation needs to be reduce if 99% of the rods must be within the tolerance?
I can not fin this question.

PROBLEM 2:
In a photographic process, the developping time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second.
- For which value is the probability 0.95 that it will be exceeded by the time it takes to develop one of the prints?
I dont get this one.

Can I have some help please?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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Well, for the second half of problem one, substitute all the values in except those given to you:

[tex] P(5.9<Z<6.1) = \phi (\frac{x_1 - \mu}{\sigma}) - \phi (\frac{x_2 - \mu}{\sigma}) [/tex]
[tex] 0.99 = \phi (\frac{6.1 - 6}{\sigma}) - \phi (\frac{5.9 - 6}{\sigma}) [/tex]
[tex] 0.99 = \phi (\frac{0.1}{\sigma}) + \phi (\frac{-0.1}{\sigma})[/tex]
[tex] 0.99 = \phi (\frac{0.1}{\sigma}) + \phi (\frac{0.1}{\sigma}) - 1[/tex]
[tex] 1.99 = 2 \phi (\frac{0.1}{\sigma}) [/tex]
[tex] 0.995 = \phi (\frac{0.1}{\sigma}) [/tex]

Use the inverse normal distribution function on 0.995, and the answer should be apparent.
 
Last edited:
  • #3
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I got the first problem.
Thanks lapo3399
 
  • #4
HallsofIvy
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Hi, I have 2 problems I would like some help. It is about normal distribution(probability)

PROBLEM 1: Extruded plastic rods are automatically cut into lenghts of 6 inches. Actual lengths are normally distributed about a mean of 6 inches and their standard deviation is 0.06 inch.

1- what proportion of the rods have lenghts that are outside the tolerance limits of 5.9 and 6.1 inches?


Here I did:
p=F((6.1-6)/0.06)- F((5.9-6)/0.06)= F(1.67)-F(-1.67)=0.9525-0.0475=0.905

P(outside tolerance)=1-0.905=0.095

2- To what value does the standard deviation needs to be reduce if 99% of the rods must be within the tolerance?
I can not fin this question.

PROBLEM 2:
In a photographic process, the developping time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second.
- For which value is the probability 0.95 that it will be exceeded by the time it takes to develop one of the prints?
I dont get this one.
In other words, Find x so that P(X> x)= 0.95. Look up the z that gives P(z)= 0.95 in the standard normal distribution and solve (x- 16.29)/0.12= z.
 

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