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

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asked 2021-05-21
The pmf of the amount of memory X (GB) in a purchased flash drive is given as the following.
\(\begin{array}{|c|c|}\hline x & 1 & 2 & 4 & 8 & 16 \\ \hline p(x) & 0.05 & 0.10 & 0.30 & 0.45 & 0.10 \\ \hline \end{array} \)
a) Compute E(X). (Enter your answer to two decimal places.) GB
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.) \(GB^{2}\)

Answers (1)

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.
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