# If random variable X has a probability density function of f(x)=1/x on the interval[e,e^2], what is the standard deviation of X?

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To find the standard deviation of a continuous random variable with a probability density function ( f(x) = \frac{1}{x} ) on the interval ( [e, e^2] ), we first need to find the mean, denoted by ( \mu ), and then use it to calculate the standard deviation.

The mean ( \mu ) of a continuous random variable is given by the integral of ( x ) times the probability density function ( f(x) ) over the interval ( [e, e^2] ), divided by the integral of ( f(x) ) over the same interval:

[ \mu = \frac{\int_{e}^{e^2} x \cdot \frac{1}{x} dx}{\int_{e}^{e^2} \frac{1}{x} dx} ]

[ = \frac{\int_{e}^{e^2} dx}{\int_{e}^{e^2} \frac{1}{x} dx} ]

[ = \frac{e^2 - e}{\ln(e^2) - \ln(e)} ]

[ = \frac{e^2 - e}{2\ln(e) - \ln(e)} ]

[ = \frac{e^2 - e}{2} ]

Now, the standard deviation ( \sigma ) can be calculated using the formula:

[ \sigma = \sqrt{\int_{e}^{e^2} (x - \mu)^2 \cdot \frac{1}{x} dx} ]

[ = \sqrt{\int_{e}^{e^2} \left(x - \frac{e^2 - e}{2}\right)^2 \cdot \frac{1}{x} dx} ]

This integral will need to be evaluated, but first, we need to simplify the expression inside the integral by expanding ( (x - \frac{e^2 - e}{2})^2 ). Once we simplify and integrate, we can then take the square root to find ( \sigma ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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