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Suppose the amount of heating oil used annually by households is normally distributed, with a mean of 760 liters per household per year and a standard deviation of 150 liters of heating oil per year.

If the members of a particular household were scard into using a fue conservation measures by newspaper acounts of the probably price of heating oil next year, and they decide they wanted to use less oil than 97.5% of all households, what is maximum amount of oil they can use and stil accomplish their objective?

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The normal distribution is defined as

$$P(X)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(X-\mu)^{2}}{2\sigma^{2}}}$$

Where X is the amount of oil for wich you want the probability, $\mu$ is the mean and $\sigma$ is the standard deviation.

What you want is to take the 97.5% of companies that use the most amount of oil, and find out how much oil is used by the most economic of those, therefore

$$\int_{X_1}^{\infty}P(X)dX=\int_{X_1}^{\infty}\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(X-\mu)^{2}}{2\sigma^{2}}}dX=0.975$$

and solve the equation for $X_1$. The logic behind this, is that this integral will give you the probability that the amount of oil used will be less than that used by 97.5% of the companies.

This integral I believe can only be calculated numerically, so making the integration start at infinity in the program and progressing until it reaches the value of 0.975 is probably the way to go, I don't know if there are more efficient methods of solving this.

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