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.

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.