Question

# The pmf of the amount of memory X (GB) in a purchased flash drive is given as the following. \begin{array}{|c|c|}\hline x & 1 & 2 & 4 & 8 & 16 \\ \hli

Decimals
The pmf of the amount of memory X (GB) in a purchased flash drive is given as the following.
$$\begin{array}{|c|c|}\hline x & 1 & 2 & 4 & 8 & 16 \\ \hline p(x) & 0.05 & 0.10 & 0.30 & 0.45 & 0.10 \\ \hline \end{array}$$
b) Compute V(X) directly from the definition. (Enter your answer to four decimal places.) $$GB^{2}$$
c) Compute the standard deviation of X. (Round your answer to three decimal places.) GB
d) Compute V(X) using the shortcut formula. (Enter your answer to four decimal places.) $$GB^{2}$$

2021-05-22
Step 1
The expected value of X is obtained below:
From the given information, the random variable X represents the amount of memory in a purchased flash drive and it takes the values 1, 2, 4, 8 and 16.
Consider,
$$E(X)=\sum_{x=1,2,4,8,16}xP_{X}(x)$$
$$=(1\times P((x=1))+(2\times P(x=2))+(4\times P(x=4))+(8\times P(x=8))+(16\times P(x=16))$$
$$=(1\times0.05)+(2\times0.10)+(4\times0.3)+(8\times0.45)+(16\times0.10)$$
$$=0.05+0.2+1.2+3.6+1.6$$
$$=6.65$$
The expected value of random variable X is obtained by summation of product of values of x and the corresponding probabilities.
Step 2
The variance of random variable X is,
$$V(X)=E(X^{2})-[E(X)]^{2}$$
Consider,
$$E(X^{2})=\sum x^{2}PX(x)$$
$$=\begin{bmatrix} (1^{2}\times P(x=1))+(2^{2}\times P(x=2))+(4^{2}\times P(x=4))+ \\ (8^{2}\times P(x=8))+(16^{2}\times P(x=16)) \end{bmatrix}$$
$$=(1^{2}\times0.05)+(2^{2}\times0.10)+(4^{2}\times0.3)+(8^{2}\times0.45)+(16^{2}\times0.10)$$
$$=0.05+0.4+4.8+28.8+25.6$$
$$=59.65$$
Thus, the value of $$E(X^{2})$$ is 59.65
$$V(X)=E(X^{2})-[E(X)]^{2}$$
$$=59.65-(6.65)^{2}$$
$$=15.4275$$
The variance of the random variable X is obtained by subtracting the square of expected value of X from the expected value of $$X^{2}$$ The variance represents the square of the spread of the amount of memory in a purchased flash drive around the mean.
Step 3
The standard deviation of random variable is,
$$\sigma=\sqrt{V(X)}$$
$$=\sqrt{15.4275}$$
$$=3.9278$$
The standard deviation of the random variable X is obtained by taking square root of the variance. The spread of the amount of memory in a purchased flash drive is 3.9278 times around the mean.
Step 4
The variance of random variable X is,
$$V(X)=E[(X-E(X))^{2}]$$
$$=\sum_{x}[(X-E(X))^{2}\times P(X=x)]$$
$$=\sum_{x}[(X-6.65)^{2}\times P(X=x)]$$
$$\left\{\begin{matrix} [(1-6.65)^{2}\times P(X=1)]+[(2-6.65)^{2}\times P(X=2)] \\ +[(4-6.65)^{2}\times P(X=4)] \\ +[(8-6.65)^{2}\times P(X=8)]+[(16-6.65)^{2}\times P(X=16)] \end{matrix}\right\}$$
$$\left\{\begin{matrix}[(1-6.65)^{2}\times0.05]+[(2-6.65)^{2}\times0.1]+[(4-6.65)^{2}\times0.3] \\ +[(8-6.65)^{2}\times0.45]+[(16-6.65)^{2}\times0.10] \end{matrix}\right\}$$
$$=1.5961+2.1623+2.1068+0.8203+8.7423$$
$$=15.4278$$
The variance of the random variable X is obtained by the expected value of square of difference between the random variable X and expected value of X. The square of the spread of the amount of memory in a purchased flash drive is 15.4275 times around the mean.