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

Step 1 
Below is how the expected value of X was determined:
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, 
$E(X)=\sum _{x=1,2,4,8,16}x{P}_X(x)$ 
$=(1\times P((x=1))+(2\times P(x=2))+(4\times P(x=4))+(8\times P(x=8))+(16\times P(x=16))$ 
$=(1\times 0.05)+(2\times 0.10)+(4\times 0.3)+(8\times 0.45)+(16\times 0.10)$ 
$=0.05+0.2+1.2+3.6+1.6$ 
$=6.65$ 
The product of the values of x and the corresponding probabilities yields the expected value of random variable X.
Step 2 
The variance of random variable X is, 
$V(X)=E(X^2)-[E(X){]}^2$ 
Consider, 
$E(X^2)=\sum x^{2}PX(x)$ 
$=\left[\begin{array}{c}(1^2\times P(x=1))+(2^2\times P(x=2))+(4^2\times P(x=4))+\\ (8^2\times P(x=8))+({16}^2\times P(x=16))\end{array}\right]$ 
$=(1^2\times 0.05)+(2^2\times 0.10)+(4^2\times 0.3)+(8^2\times 0.45)+({16}^2\times 0.10)$ 
$=0.05+0.4+4.8+28.8+25.6$ 
$=59.65$ 
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, 
$V(X)=E[(X-E(X){)}^2]$ 
$=\sum _x[(X-E(X){)}^2\times P(X=x)]$ 
$=\sum _x[(X-6.65{)}^2\times P(X=x)]$ 
$\left\{\begin{array}{c}[(1-6.65{)}^2\times P(X=1)]+[(2-6.65{)}^2\times P(X=2)]\\ +[(4-6.65{)}^2\times P(X=4)]\\ +[(8-6.65{)}^2\times P(X=8)]+[(16-6.65{)}^2\times P(X=16)]\end{array}\right\}$ 
$\left\{\begin{array}{c}[(1-6.65{)}^2\times 0.05]+[(2-6.65{)}^2\times 0.1]+[(4-6.65{)}^2\times 0.3]\\ +[(8-6.65{)}^2\times 0.45]+[(16-6.65{)}^2\times 0.10]\end{array}\right\}$