Question

1. how do you find the variance and standard deviation of the probability distribution?
2. which formula is easier in finding the variance and standard deviation of the probability distribution? elaborate why.

Explanation:

Step1: Variance formula

The variance $\sigma^{2}$ of a probability distribution is given by $\sigma^{2}=\sum(x - \mu)^{2}P(x)$ where $x$ are the values of the random variable, $\mu$ is the mean ($\mu=\sum xP(x)$) and $P(x)$ is the probability of $x$. Another formula for variance is $\sigma^{2}=\sum x^{2}P(x)-\mu^{2}$.

Step2: Standard deviation formula

The standard deviation $\sigma$ is the square - root of the variance. So if variance is $\sigma^{2}$, then $\sigma=\sqrt{\sigma^{2}}$.

Step3: Easier formula

The formula $\sigma^{2}=\sum x^{2}P(x)-\mu^{2}$ is often easier. When calculating $\sum(x - \mu)^{2}P(x)$, we first need to calculate $(x-\mu)$ for each $x$, then square it and multiply by $P(x)$ and sum. For $\sum x^{2}P(x)-\mu^{2}$, we can calculate $\sum x^{2}P(x)$ (by squaring $x$ first and then multiplying by $P(x)$ and summing) and then subtract $\mu^{2}$ (where $\mu$ is already calculated as $\sum xP(x)$).

Answer:

1. Variance: $\sigma^{2}=\sum(x - \mu)^{2}P(x)$ or $\sigma^{2}=\sum x^{2}P(x)-\mu^{2}$; Standard deviation: $\sigma=\sqrt{\sum(x - \mu)^{2}P(x)}$ or $\sigma=\sqrt{\sum x^{2}P(x)-\mu^{2}}$.
2. The formula $\sigma^{2}=\sum x^{2}P(x)-\mu^{2}$ is easier as it reduces the number of sub - calculations (less arithmetic operations involving $(x - \mu)$).