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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

The College Board reports that \(2 \%\) of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. a. What is the probability that exactly 1 received a special accommodation? b. What is the probability that at least 1 received a special accommodation? c. What is the probability that at least 2 received a special accommodation? d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed \(4.5\) hours. What would you expect the average time allowed the 25 selected students to be?

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Planning and Mgmnt., 2005: 383-393), the drought length \(Y\) is the number of consecutive time intervals in which the water supply remains below a critical value \(y_{0}\) (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). The cited paper proposes a geometric distribution with \(p=.409\) for this random variable. a. What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? b. What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. a. What is the pmf of the number of granite specimens selected for analysis? b. What is the probability that all specimens of one of the two types of rock are selected for analysis? c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of five per hour. a. What is the probability that exactly four arrivals occur during a particular hour? b. What is the probability that at least four people arrive during a particular hour? c. How many people do you expect to arrive during a 45 min period?

Suppose that \(90 \%\) of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? What assumptions did you make in the course of answering the question posed?
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