Question

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

Decimals
ANSWERED
asked 2021-09-20

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.) \(\displaystyle{G}{B}^{{{2}}}\)
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.) \(\displaystyle{G}{B}^{{{2}}}\)

Expert Answers (1)

2021-09-21

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.

21
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...