Standard Deviation

Question

For the genetics of a given plant, the probability of any seed it produces being brown is 3/4. The plant produces 80 seeds. Let $X$ be the random variable that counts the number of brown seeds. Find the expected value and the standard deviation of $X$.

So... I figured the expected value of $X$ to be 80 x 3/4 = 60. But what I don't understand is how do you figure the standard deviation of $X$ when you have a brown seed or not! How do you get a mean from this?

Answer 1

Let $Y_i$ be a random variable taking the value $1$ if the seed is brown and $0$ otherwise. Moreover, I assume the all $Y_i 's$ are independent. As you already correctly stated, the expectation is

$E[X] = E[\sum_{i=1}^{80} Y_i] = \sum_{i=1}^{80} E[Y_i] = \sum_{i=1}^{80} 0.75 = 60$.

The standard deviation is defined as the square root of the variance of a random variable: $SD[X] = \sqrt{Var[X]}$. The Variance is defined as $Var[X] = E[X^2] - E[X]^2$. In your case,

$Var[X] = Var[\sum_{i=1}^{80} Y_i] = E[\sum_{i=1}^{80} Y_i^2] - E[\sum_{i=1}^{80} Y_i]^2$.

Now, since the all $Y_i$ are independent the variance of the sum is equal to the sum of variances. Moreover, you know already the expectation for one $Y_i$. It's $0.75$. In addition, $E[X^2] = E[X]$ because squaring 0 or 1 is again 0 or 1. Hence, you get

$Var[\sum_{i=1}^{80} Y_i] = \sum_{i=1}^{80} Var[ Y_i] \\ \qquad \qquad \qquad = \sum_{i=1}^{80}(E[Y_i^2] - E[Y_i]^2) \\ \qquad \qquad \qquad= \sum_{i=1}^{80} (E[Y_i] - 0.75^2) = 80(0.75 - 0.75^2) = 15 $.

So, your standard deviation is eventually $SD[X] = \sqrt{Var[Y_i]} = \sqrt{15}$