Normal Distribution question

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October 15th 2009, 06:03 PM
Oblivionwarrior

Normal Distribution question

I am stuck on this question, I don't exactly know how to approach it. Thanks for any advice.

The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed. Suppose that the mean of the filling operation can be adjusted easily, but the standard deviation remains at 0.1 ounce.

a) At what value should the mean be set so that 99.9% of all cans exceed 12 ounces?

b) At what value should the mean be set so that 99.9% of all cans exceed 12 ounces if the standard deviation can be reduced to 0.05 fluid ounce?
October 15th 2009, 07:47 PM
mr fantastic

Quote:

Originally Posted by Oblivionwarrior

I am stuck on this question, I don't exactly know how to approach it. Thanks for any advice.

The fill volume of an automated filling machine used for filling cans of carbonated beverage is normally distributed. Suppose that the mean of the filling operation can be adjusted easily, but the standard deviation remains at 0.1 ounce.

a) At what value should the mean be set so that 99.9% of all cans exceed 12 ounces?

b) At what value should the mean be set so that 99.9% of all cans exceed 12 ounces if the standard deviation can be reduced to 0.05 fluid ounce?

Here is (a). Think about (b).

You know that $\Pr(X < 12) = 0.01$ . Get (using tables or however you've been taught) the value of z* such that $\Pr(Z < z^*) < 0.01$ .

Then $Z = \frac{X - \mu}{\sigma} \Rightarrow z^* = \frac{12 - \mu}{0.1}$ . Solve for $\mu$ .
October 16th 2009, 09:12 AM
Oblivionwarrior

Thanks. I have another quick question. It isn't related to one posted earlier. I want to know if I am doing this correctly.

Assume X is normally distributed with a mean of 13 and a standard deviation of 2. Determine the value for x that solves each of the following

If the question asks P(X > x) = 0.95

My steps(not sure if they are correct)

P(X > x) = 0.95
1 - P(X < x) = 0.95
1 - P(Z < (x - 13)/2) = 0.95
P(Z < (x - 13)/2) = 0.05

now, looking at the z-chart to find the value which corresponds to the probability 0.05 I find it to be approximately -1.64. So,

(x - 13)/2 = -1.64, x = 9.72

Are these steps correct? Or is something wrong? Thanks.
October 16th 2009, 10:08 AM
aman_cc

Quote:

Originally Posted by Oblivionwarrior

Thanks. I have another quick question. It isn't related to one posted earlier. I want to know if I am doing this correctly.

Assume X is normally distributed with a mean of 13 and a standard deviation of 2. Determine the value for x that solves each of the following

If the question asks P(X > x) = 0.95

My steps(not sure if they are correct)

P(X > x) = 0.95
1 - P(X < x) = 0.95
1 - P(Z < (x - 13)/2) = 0.95
P(Z < (x - 13)/2) = 0.05

now, looking at the z-chart to find the value which corresponds to the probability 0.05 I find it to be approximately -1.64. So,

(x - 13)/2 = -1.64, x = 9.72

Are these steps correct? Or is something wrong? Thanks.

correct !
October 16th 2009, 04:19 PM
Oblivionwarrior

I have one more question regarding the one I posted above. How exactly would I go about finding x for this question?

P(-x < X - 13 < x) = 0.95

would I add 13 to both sides so its like P(-x + 13 < X < x + 13)?
October 16th 2009, 08:29 PM
mr fantastic

Quote:

Originally Posted by Oblivionwarrior

I have one more question regarding the one I posted above. How exactly would I go about finding x for this question?

P(-x < X - 13 < x) = 0.95

would I add 13 to both sides so its like P(-x + 13 < X < x + 13)?

Yes, that's one way of doing it. Now note that $\Pr(X > x + 13) = 0.025$ So find a value $z^*$ of $Z$ such that $\Pr(Z > z^*) = 0.025$ and then solve $z^* = \frac{(x + 13) - 13}{2}$ .