# 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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