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Chapter 3: Problem 29

The pmf of the amount of memory \(X(G B)\) in a purchased flash drive was given in Example \(3.13\) as \begin{tabular}{l|lllll} \(x\) & 1 & 2 & 4 & 8 & 16 \\ \hline\(p(x)\) & \(.05\) & \(.10\) & \(.35\) & \(.40\) & \(.10\) \end{tabular} 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

Short Answer

Expert verified
a. \(E(X) = 6.45\) b. \(V(X) = 15.64\) c. Standard deviation \(\approx 3.95\) d. \(V(X) \approx 15.65\)

Step by step solution

01

Understanding the PMF

The probability mass function (PMF) gives us the probabilities for discrete random variable values. For this problem, the different memory sizes and their probabilities are given: - \(X = 1\) GB with \(p(1) = 0.05\)- \(X = 2\) GB with \(p(2) = 0.10\)- \(X = 4\) GB with \(p(4) = 0.35\)- \(X = 8\) GB with \(p(8) = 0.40\)- \(X = 16\) GB with \(p(16) = 0.10\)
02

Calculating Expected Value (E(X))

The expected value \(E(X)\) is calculated using the formula \(E(X) = \sum x_i p(x_i)\). Compute it:\[E(X) = 1 \times 0.05 + 2 \times 0.10 + 4 \times 0.35 + 8 \times 0.40 + 16 \times 0.10\]\[E(X) = 0.05 + 0.20 + 1.4 + 3.2 + 1.6 = 6.45\]
03

Calculating Variance (V(X)) from Definition

Variance is calculated using the formula \(V(X) = \sum (x_i - E(X))^2 p(x_i)\). Compute each component first:- For \(x = 1\): \((1 - 6.45)^2 \times 0.05 = 0.05 \times 29.7025 = 1.485125\)- For \(x = 2\): \((2 - 6.45)^2 \times 0.10 = 0.10 \times 19.8025 = 1.98025\)- For \(x = 4\): \((4 - 6.45)^2 \times 0.35 = 0.35 \times 6.0025 = 2.100875\)- For \(x = 8\): \((8 - 6.45)^2 \times 0.40 = 0.40 \times 2.4025 = 0.961\)- For \(x = 16\): \((16 - 6.45)^2 \times 0.10 = 0.10 \times 91.1025 = 9.11025\)Adding these gives:\[V(X) = 1.485125 + 1.98025 + 2.100875 + 0.961 + 9.11025 = 15.64\]
04

Calculating the Standard Deviation of X

The standard deviation is the square root of the variance.\[\text{Standard Deviation} = \sqrt{V(X)} = \sqrt{15.64} \approx 3.95\]
05

Calculating Variance (V(X)) using the Shortcut Formula

The shortcut formula for variance is \(V(X) = E(X^2) - (E(X))^2\).First calculate \(E(X^2)\):\[E(X^2) = 1^2 \times 0.05 + 2^2 \times 0.10 + 4^2 \times 0.35 + 8^2 \times 0.40 + 16^2 \times 0.10\]\[E(X^2) = 0.05 + 0.40 + 5.6 + 25.6 + 25.6 = 57.25\]Now use the shortcut formula:\[V(X) = 57.25 - (6.45)^2 = 57.25 - 41.6025 = 15.6475\]Note this value is close to the variance calculated directly due to rounding differences.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value is a key concept in probability distribution, representing the average outcome you would expect if you repeated an experiment many times. This is not merely an arithmetic mean; it is a weighted average where each possible outcome is multiplied by its respective probability.To calculate it, use the formula: \[E(X) = \sum x_i p(x_i)\]
  • This means you multiply each possible value of the random variable, \(x\), by the probability of that value occurring, \(p(x)\), and then sum all these products.
  • In the example provided, the expected value for sizes of memory in a flash drive is calculated to be 6.45GB.
This tells you, statistically, the average memory size you'd expect to buy if you randomly purchased a large number of drives, given the provided probabilities.
Variance
Variance measures how much the values in a probability distribution differ from the expected value.It provides insight into the spread or dispersion of data around the mean, with higher variance indicating more spread out data points.Using the formula for variance: \[ V(X) = \sum (x_i - E(X))^2 p(x_i) \]
  • You subtract the expected value from each possible value of the random variable, square it to remove negatives, multiply by the probability of the value, and sum all these products for each possible outcome.
  • In this situation, when calculated directly, the variance for the flash drive memory sizes was found to be 15.64.
This shows that the memory sizes purchased vary quite a bit from the average of 6.45GB.
Standard Deviation
The standard deviation is closely related to variance, providing a more interpretable measure of spread in the same units as the random variable.It is the square root of the variance, which 'reverses' the squaring step in variance calculations, and is denoted as:\[\text{Standard Deviation} = \sqrt{V(X)}\]
  • Because it is derived from variance, the standard deviation will also reflect how spread out the values are relative to the expected value.
  • In our exercise, with a calculated variance of 15.64, the standard deviation was approximately 3.95GB.
This means that the memory size of the flash drives deviates from the average by about 3.95GB on average, portraying the variability in purchases.
Probability Mass Function
A probability mass function (PMF) gives the probability that a discrete random variable is exactly equal to some value. It is specific to discrete variables and assigns probabilities to each possible outcome.PMF is represented as:\[p(x) = P(X = x)\]
  • In the given problem, the PMF describes the probability distribution of the different flash drive memory sizes available for purchase.
  • This PMF allowed us to calculate the expected value, variance, and standard deviation.
  • Here, each memory size鈥�1GB, 2GB, 4GB, 8GB, and 16GB鈥攈ad a specific probability, such as 0.05 for 1GB, explained via the PMF.
The PMF forms the foundational tool for carrying out these calculations by providing the probabilities needed for analysis.

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Most popular questions from this chapter

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5 -lb batches. Let \(X=\) the number of batches ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X\).]

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of boards 1 and 2 for inspection. a. List the ten different possible outcomes. b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(X\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(X\). c. Let \(F(x)\) denote the odf of \(X\). First determine \(F(0)=\) \(P(X \leq 0), F(1)\), and \(F(2)\); then obtain \(F(x)\) for all other \(x\).

If the sample space \(S\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(\&\) will have an infinite set of possible values? If yes, say why. If no, give an example.

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, \(25 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. Then one possible outcome is SFSS, with probability \((.25)(.75)(.25)(.25)\) and associated \(X\) value 3 . There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?

A toll bridge charges \(\$ 1.00\) for passenger cars and \(\$ 2.50\) for other vehicles. Suppose that during daytime hours, \(60 \%\) of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? [Hint: Let \(X=\) the number of passenger cars; then the toll revenue \(h(X)\) is a linear function of \(\left.X_{-}\right]\)
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