# Value of a normal random variable.

• Jun 17th 2009, 10:27 PM
shinn
Value of a normal random variable.
I just got another quick question related to this.

"Nominal 375mL cans of soft drink have a mean content slightly above this value so that the stated contents are exceeded most of the time. Suppose the contents are normally distributed with standard deviation of 5mL.

Find the mean content so that only 1% of the cans are under the stated value"

This is what I've done so far, but comparing to the answers it is incorrect:

Let $X_i$ be the i'th measurement of the nominal cans for i = 1,2,3...,k

So, Average $X_k$ ~ $N(\mu , 5^2)$
Let $\mu$ = 375 + d, for d>0.

We need to find P( Average $X_k$ < 375) = 0.01

$LHS = P( z < (375 - \mu) / 5)$

Simplifying and using tables, d / 5 = .8389. Thus, d = 4.195.

And so the mean content = 375 + 4.195 = 379.2 mL.

However, the answers say the mean content is 387 mL.

Once again, any help will be really appreciated (Nod)
• Jun 17th 2009, 11:31 PM
mr fantastic
Quote:

Originally Posted by shinn
I just got another quick question related to this.

"Nominal 375mL cans of soft drink have a mean content slightly above this value so that the stated contents are exceeded most of the time. Suppose the contents are normally distributed with standard deviation of 5mL.

Find the mean content so that only 1% of the cans are under the stated value"

This is what I've done so far, but comparing to the answers it is incorrect:

Let $X_i$ be the i'th measurement of the nominal cans for i = 1,2,3...,k

So, Average $X_k$ ~ $N(\mu , 5^2)$
Let $\mu$ = 375 + d, for d>0.

We need to find P( Average $X_k$ < 375) = 0.01

$LHS = P( z < (375 - \mu) / 5)$

Simplifying and using tables, d / 5 = .8389. Thus, d = 4.195.

And so the mean content = 375 + 4.195 = 379.2 mL.

However, the answers say the mean content is 387 mL.

Once again, any help will be really appreciated (Nod)

$Z = \frac{X - \mu}{\sigma} = \frac{X - 375}{5}$.

Substitute z = z* where Pr(Z < z*) = 0.99 and solve for X.