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,

\(\displaystyle{E}{\left({X}\right)}=\sum_{{{x}={1},{2},{4},{8},{16}}}{x}{P}_{{{X}}}{\left({x}\right)}\)

\(\displaystyle={\left({1}\times{P}{\left({\left({x}={1}\right)}\right)}+{\left({2}\times{P}{\left({x}={2}\right)}\right)}+{\left({4}\times{P}{\left({x}={4}\right)}\right)}+{\left({8}\times{P}{\left({x}={8}\right)}\right)}+{\left({16}\times{P}{\left({x}={16}\right)}\right)}\right.}\)

\(\displaystyle={\left({1}\times{0.05}\right)}+{\left({2}\times{0.10}\right)}+{\left({4}\times{0.3}\right)}+{\left({8}\times{0.45}\right)}+{\left({16}\times{0.10}\right)}\)

\(\displaystyle={0.05}+{0.2}+{1.2}+{3.6}+{1.6}\)

\(\displaystyle={6.65}\)

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,

\(\displaystyle{E}{\left({X}^{{{2}}}\right)}=\sum{x}^{{{2}}}{P}{X}{\left({x}\right)}\)

\(=\begin{bmatrix} (1^{2}\times P(x=1))+(2^{2}\times P(x=2))+(4^{2}\times P(x=4))+ \\ (8^{2}\times P(x=8))+(16^{2}\times P(x=16)) \end{bmatrix}\)

\(\displaystyle={\left({1}^{{{2}}}\times{0.05}\right)}+{\left({2}^{{{2}}}\times{0.10}\right)}+{\left({4}^{{{2}}}\times{0.3}\right)}+{\left({8}^{{{2}}}\times{0.45}\right)}+{\left({16}^{{{2}}}\times{0.10}\right)}\)

\(\displaystyle={0.05}+{0.4}+{4.8}+{28.8}+{25.6}\)

\(\displaystyle={59.65}\)

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,

\(\displaystyle{V}{\left({X}\right)}={E}{\left[{\left({X}-{E}{\left({X}\right)}\right)}^{{{2}}}\right]}\)

\(\displaystyle=\sum_{{{x}}}{\left[{\left({X}-{E}{\left({X}\right)}\right)}^{{{2}}}\times{P}{\left({X}={x}\right)}\right]}\)

\(\displaystyle=\sum_{{{x}}}{\left[{\left({X}-{6.65}\right)}^{{{2}}}\times{P}{\left({X}={x}\right)}\right]}\)

\(\left\{\begin{matrix} [(1-6.65)^{2}\times P(X=1)]+[(2-6.65)^{2}\times P(X=2)] \\ +[(4-6.65)^{2}\times P(X=4)] \\ +[(8-6.65)^{2}\times P(X=8)]+[(16-6.65)^{2}\times P(X=16)] \end{matrix}\right\}\)

\(\left\{\begin{matrix}[(1-6.65)^{2}\times0.05]+[(2-6.65)^{2}\times0.1]+[(4-6.65)^{2}\times0.3] \\ +[(8-6.65)^{2}\times0.45]+[(16-6.65)^{2}\times0.10] \end{matrix}\right\}\)

\(\displaystyle={1.5961}+{2.1623}+{2.1068}+{0.8203}+{8.7423}\)

\(\displaystyle={15.4278}\)

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.