Normal distribution of rods

[brad sue]

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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 don't get this one.
Can I have some help please?

 

[lapo3399]

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

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

Use the inverse normal distribution function on 0.995, and the answer should be apparent.

 

[brad sue]

I got the first problem.
Thanks lapo3399

 

[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 don't 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.