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The length of time it takes to fill an order at a local sandwich shop is normally distributed with a mean of 4.1 minutes and standard deviation of 1.3 minutes. What is the probability that you have to wait more than 6 minutes for a sandwich?

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Let random variable $W$ be the waiting time. Then $$\Pr(W\gt 6)=\Pr(W-4.1\gt 6-4.1)=\Pr\left(\frac{W-4.1}{1.3}\gt \frac{6-4.1}{1.3}\right).$$

But the random variable $\frac{W-4.1}{1.3}$ has standard normal distribution. So you want to find $$\Pr\left( Z\gt \frac{6-4.1}{1.3}\right),$$ where $Z$ has standard normal distribution.

Now you need to calculate $\frac{6-4.1}{1.3}$, and use a table of the standard normal, or appropriate software.

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$Pr(T>6)=1-Pr(T\leq 6) =1-\int_{-\infty}^6 \frac{1}{1.3 (2 pi)^2} e^{-\frac{((x-4.1)/1.3)^2}{2} } dx =0.0719339$

all i used is in here: http://www.wolframalpha.com/input/?i=1-int+1%2F%281.3+%282+pi%29^%281%2F2%29%29+e^%28-%28%28x-4.1%29%2F1.3%29^2+%2F2%29+dx+from+-infty+to+6

http://en.wikipedia.org/wiki/Normal_distribution

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