The pmf of the amount of memory X (GB) in a purchased flash drive is given as th

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,
${\displaystyle E\left(X\right)=\sum _{x=1,2,4,8,16}x{P}_X\left(x\right)}$
${\displaystyle =(1\times P\left((x=1)\right)+(2\times P(x=2))+(4\times P(x=4))+(8\times P(x=8))+(16\times P(x=16))}$
${\displaystyle =(1\times 0.05)+(2\times 0.10)+(4\times 0.3)+(8\times 0.45)+(16\times 0.10)}$
${\displaystyle =0.05+0.2+1.2+3.6+1.6}$
${\displaystyle =6.65}$
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,
${\displaystyle V\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2}$
Consider,
${\displaystyle E\left(X^2\right)=\sum x^{2}PX\left(x\right)}$
$=\left[\begin{array}{c}(1^2\times P(x=1))+(2^2\times P(x=2))+(4^2\times P(x=4))+\\ (8^2\times P(x=8))+({16}^2\times P(x=16))\end{array}\right]$
${\displaystyle =(1^2\times 0.05)+(2^2\times 0.10)+(4^2\times 0.3)+(8^2\times 0.45)+({16}^2\times 0.10)}$
${\displaystyle =0.05+0.4+4.8+28.8+25.6}$
${\displaystyle =59.65}$
Thus, the value of ${\displaystyle E\left(X^2\right)}$ is 59.65
${\displaystyle V\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2}$
${\displaystyle =59.65-{\left(6.65\right)}^2}$
${\displaystyle =15.4275}$
The variance of the random variable X is obtained by subtracting the square of expected value of X from the expected value of ${\displaystyle X^2}$ 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,
${\displaystyle \sigma =\sqrt{V\left(X\right)}}$
${\displaystyle =\sqrt{15.4275}}$
${\displaystyle =3.9278}$
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,
${\displaystyle V\left(X\right)=E\left[{(X-E\left(X\right))}^2\right]}$
${\displaystyle =\sum _x[{(X-E\left(X\right))}^2\times P(X=x)]}$