Question

If x is a binomial random variable, compute the mean, the standard deviation, and the variance for each of the following cases:

Answer 1

Answer:

(a) $\mu_X=1.2$, $\sigma_X=0.8485$, $\sigma^{2} _X=0.72$

(b) $\mu_X=4$, $\sigma_X=0.8944$, $\sigma^{2} _X=0.8$

(c) $\mu_X=1.5$, $\sigma_X=1.0247$, $\sigma^{2} _X=1.05$

(d) $\mu_X=3.2$, $\sigma_X=0.8$, $\sigma^{2} _X=0.64$

Step-by-step explanation:

A binomial random variable counts how often a particular event occurs in a fixed number of tries or trials.

For a binomial random variable $X$, we can use these formulas for mean, standard deviation, and variance:

$\mu_X=np$

$\sigma_X=\sqrt{np(1-p)}$

$\sigma^{2} _X=np(1-p)$

Applying the above formulas we get,

(a) n = 3, p = 0.4

$\mu_X=3\cdot 0.4=1.2$

$\sigma_X=\sqrt{3\cdot 0.4(1-0.4)}=0.8485$

$\sigma^{2} _X=3\cdot \:0.4\left(1-0.4\right)=0.72$

(b) n = 5, p = 0.8

$\mu_X=5\cdot 0.8=4$

$\sigma_X=\sqrt{5\cdot 0.8(1-0.8)}=0.8944$

$\sigma^{2} _X=5\cdot \:0.8\left(1-0.8\right)=0.8$

(c) n = 5, p = 0.3

$\mu_X=5\cdot 0.3=1.5$

$\sigma_X=\sqrt{5\cdot 0.3(1-0.3)}=1.0247$

$\sigma^{2} _X=5\cdot \:0.3\left(1-0.3\right)=1.05$

(d) n = 4, p = 0.8

$\mu_X=4\cdot 0.8=3.2$

$\sigma_X=\sqrt{4\cdot 0.8(1-0.8)}=0.8$

$\sigma^{2} _X=4\cdot \:0.8\left(1-0.8\right)=0.64$