Originally Posted by

**icemanfan** Ok, so you want the probability that more than 300 meals are served given that the mean is 250 and the standard deviation is 37. The probability density function for the normal distribution is

$\displaystyle \frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\frac{(x - \mu)^2}{2\sigma^2}\right)$.

Therefore, the answer is

$\displaystyle \int _{300} ^{\infty} \frac{1}{37\sqrt{2\pi}}\exp\left(-\frac{(x - 250)^2}{2 \cdot 37^2}\right)dx$.

Alternatively, you notice that 300 is 1.35 standard deviations away from 250 and use the table.