Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 29E from Chapter 3 from Devore's Probability and Statistics for Engineering and Science.
Problem 29E
Chapter:
Problem:
The pmf of the amount of memory X (GB) in a purchased flash drive was given in Example 3.13 as...Compute the following: a. E(X) b. V(X) directly from the definition c. The standard deviation of X d. V(X) using the shortcut formula
Step-by-Step Solution
Step 1
We are given with probability distribution function:\[p\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{0.05}&{x = 1}\\{0.10}&{x = 2}\\{0.35}&{x = 4}\\{0.40}&{x = 8}\\{0.10}&{x = 16}\end{array}} \right.\]
Step 2: (a) E(x)
Expected value of x \[\begin{array}{l}E\left( x \right) = \Sigma xp\left( x \right)\\ = \left( {1 \times 0.05} \right) + \left( {2 \times 0.1} \right) + \left( {4 \times 0.35} \right) + \left( {8 \times 0.4} \right) + \left( {16 \times 0.1} \right)\\ = 0.05 + 0.2 + 1.4 + 3.2 + 1.6\\ = 6.45\end{array}\]Therefore, \[E\left( x \right) = 6.45\]
Step 3: (b)
Variance of X\[\begin{array}{l}V\left( x \right) = \sum {{{\left( {x - E\left( x \right)} \right)}^2}p\left( x \right)} \\ = {\left( {1 - 6.45} \right)^2}p\left( 1 \right) + {\left( {2 - 6.45} \right)^2}p\left( 2 \right) + {\left( {4 - 6.45} \right)^2}p\left( 4 \right) + {\left( {8 - 6.45} \right)^2}p\left( 8 \right) + {\left( {16 - 6.45} \right)^2}p\left( {16} \right)\\ = {\left( {1 - 6.45} \right)^2}\left( {0.05} \right) + {\left( {2 - 6.45} \right)^2}\left( {0.1} \right) + {\left( {4 - 6.45} \right)^2}\left( {0.35} \right) + {\left( {8 - 6.45} \right)^2}\left( {0.4} \right) + {\left( {16 - 6.45} \right)^2}\left( {0.1} \right)\\ = 1.485125 + 1.98025 + 2.100875 + 0.961 + 9.12025\\ = 15.6475\end{array}\]Therefore, \[V\left( x \right) = 15.6475\]
Step 4: (c)
The standard deviation of X is given by square root of the variance calculated in above step:\[\begin{array}{l}SD\left( x \right) = \sqrt {V\left( x \right)} \\ = \sqrt {15.6475} \\ = 3.9556921\\ \approx 3.9557\end{array}\]Therefore, \[SD\left( x \right) \approx 3.9557\]
We are given with probability distribution function:\[p\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{0.05}&{x = 1}\\{0.10}&{x = 2}\\{0.35}&{x = 4}\\{0.40}&{x = 8}\\{0.10}&{x = 16}\end{array}} \right.\]
Step 2: (a) E(x)
Expected value of x \[\begin{array}{l}E\left( x \right) = \Sigma xp\left( x \right)\\ = \left( {1 \times 0.05} \right) + \left( {2 \times 0.1} \right) + \left( {4 \times 0.35} \right) + \left( {8 \times 0.4} \right) + \left( {16 \times 0.1} \right)\\ = 0.05 + 0.2 + 1.4 + 3.2 + 1.6\\ = 6.45\end{array}\]Therefore, \[E\left( x \right) = 6.45\]
Step 3: (b)
Variance of X\[\begin{array}{l}V\left( x \right) = \sum {{{\left( {x - E\left( x \right)} \right)}^2}p\left( x \right)} \\ = {\left( {1 - 6.45} \right)^2}p\left( 1 \right) + {\left( {2 - 6.45} \right)^2}p\left( 2 \right) + {\left( {4 - 6.45} \right)^2}p\left( 4 \right) + {\left( {8 - 6.45} \right)^2}p\left( 8 \right) + {\left( {16 - 6.45} \right)^2}p\left( {16} \right)\\ = {\left( {1 - 6.45} \right)^2}\left( {0.05} \right) + {\left( {2 - 6.45} \right)^2}\left( {0.1} \right) + {\left( {4 - 6.45} \right)^2}\left( {0.35} \right) + {\left( {8 - 6.45} \right)^2}\left( {0.4} \right) + {\left( {16 - 6.45} \right)^2}\left( {0.1} \right)\\ = 1.485125 + 1.98025 + 2.100875 + 0.961 + 9.12025\\ = 15.6475\end{array}\]Therefore, \[V\left( x \right) = 15.6475\]
Step 4: (c)
The standard deviation of X is given by square root of the variance calculated in above step:\[\begin{array}{l}SD\left( x \right) = \sqrt {V\left( x \right)} \\ = \sqrt {15.6475} \\ = 3.9556921\\ \approx 3.9557\end{array}\]Therefore, \[SD\left( x \right) \approx 3.9557\]