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Suppose that instant coffee comes in two sizes, 10-ounce jars and 30-ounce jars.

Let $X$ be the actual weight of coffee in a 10-ounce jar and assume that $X$ has a normal distribution with a mean 10.1 ounces and standard deviation 0.2 ounces.

Let $Y$ be the actual weight of coffee in a 30-ounce jar and assume that $Y$ has a normal distribution with mean 30.4 ounces and standard deviation 0.42 ounces.

Assume that the weights $X$ and $Y$ are independent random variables.

Find the probability that total weight of coffee is three 10-ounce jars in greater than the weight in one 30-ounce jar.

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Hint: What kind of random variable is the sum of independent normal random variables? the difference of independent normal random variables? How can you find the probability that the difference of two normal random variables exceeds $0$? – Dilip Sarwate Jan 25 '13 at 13:37

Let $X_1$, $X_2$ and $X_3$ be weights of the three 10-ounce jars. Assume them to be independent random variables. The question asks to evaluate: $$ \mathbb{P}\left(X_1+X_2+X_3 > Y\right) = \mathbb{P}\left(X_1+X_2+X_3 -Y > 0\right) $$ Let $Z = X_1+X_2+X_3 -Y$. Observe, that since $Z$ is a linear combination of normal random variables, $Z$ is also a normal random variable. The normal random variable is determined by its mean and variance: $$ \mu_Z = \mathbb{E}\left(Z\right) = \mathbb{E}\left(X_1+X_2+X_3 -Y\right) = 3 \mu_X - \mu_Y $$ $$ \sigma_Z^2 = \mathbb{Var}\left(Z\right) = \mathbb{Var}\left(X_1+X_2+X_3 -Y\right) = 3 \sigma_X^2 + \sigma_Y^2 $$ where the linearity of the expectation was used to find the mean, and the law of the total variation and independence of random variables was used to find $\sigma_Z^2$. It now remains to find $$ \mathbb{P}(Z > 0) = \Phi\left(-\frac{\mu_Z}{\sigma_Z}\right) $$ where $\Phi$ is the cumulative distribution function of the standard normal random variable.

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This is very helpful. Do you now use the values of the mean of x and mean of y to plug into the E(Z) and similarly square the standard deviations given to find Var(Z)? – user59633 Jan 26 '13 at 18:54

Since $X\sim \mathcal{N}(10.1,0.2^2)$ and $Y\sim N(30.4,0.42^2)$ and $Z=3X-Y$ be a linear combination of $X,Y$ . So here $Z$ follows normal distribution with mean $E(Z)=E(3X-Y)=3E(X)-E(Y)=-0.1$ and variance $Var(Z)=Var(3X-Y)=9Var(X)+Var(Y)=0.5364$. Now compute the probability $P[Z>0]$.

EDIT: Note that $Z \sim\mathcal{N}(-0.1,0.5364)$

So $$\begin{align}P[Z>0] &=P[\frac{Z+0.1}{\sqrt{0.5364}}>\frac{-0.1}{\sqrt{0.5364}}] \\ &=P[U>-.14] \\&=P[U <.14]=0.5557 \end{align}$$ where $U=\frac{Z+0.1}{\sqrt{0.5364}}\sim \mathcal{N}(0,1^2)$ and since $U$ is symmetric about $0$ and continuous we have $P[U >-a] =P[U <a] $. To calculate the value of the probability $P[U <.14]$ you need to use Bimetrica Table for Statistician .

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how do you compute $P[Z>0]$? – user59633 Jan 27 '13 at 21:30
@user59633 : See my answer. I edit it. – A.D Jan 29 '13 at 11:52
yes! Thank you! – user59633 Feb 1 '13 at 2:13

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