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Chapter 8: Problem 3

v$$ \begin{array}{lccccc} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline P(X=x) & 1 / 16 & 4 / 16 & 6 / 16 & 4 / 16 & 1 / 16 \\ \hline \end{array} $$

Short Answer

Expert verified
The given probability distribution is valid, and it has a mean \(E(X) = 0\), variance \(Var(X) = \frac{3}{2}\), and standard deviation \(SD(X) = \sqrt{\frac{3}{2}}\).

Step by step solution

01

First, we need to check if the sum of all probabilities is equal to 1 and if they are between 0 and 1. The sum of the probabilities is \(\frac{1}{16} + \frac{4}{16} + \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = \frac{16}{16} = 1\). All probabilities are also between 0 and 1, so it is a valid probability distribution. #Step 2: Find the mean (expected value) of the probability distribution#

To find the mean of the probability distribution, we use the formula \(E(X) = \sum xP(X=x)\). Multiply each value of x with its corresponding probability P(X=x) and then sum them up: \(E(X) = (-2)\left(\frac{1}{16}\right) + (-1)\left(\frac{4}{16}\right) + (0)\left(\frac{6}{16}\right) + (1)\left(\frac{4}{16}\right) + (2)\left(\frac{1}{16}\right) = -\frac{2}{16} - \frac{4}{16} + \frac{4}{16} + \frac{2}{16} = 0\) So, the mean of the probability distribution is 0. #Step 3: Find the variance of the probability distribution#
02

To find the variance of the probability distribution, we use the formula \(Var(X) = E(X^2) - [E(X)]^2\). We have already calculated E(X), so now we need to calculate E(X^2): \(E(X^2) = (-2)^2\left(\frac{1}{16}\right) + (-1)^2\left(\frac{4}{16}\right) + (0)^2\left(\frac{6}{16}\right) + (1)^2\left(\frac{4}{16}\right) + (2)^2\left(\frac{1}{16}\right) = \frac{4}{16} + \frac{4}{16} + \frac{16}{16} = \frac{24}{16}\) Now, plug E(X) and E(X^2) into the variance formula: \(Var(X) = \frac{24}{16} - (0)^2 = \frac{3}{2}\) So, the variance of the probability distribution is \(\frac{3}{2}\). #Step 4: Find the standard deviation of the probability distribution#

The standard deviation is the square root of the variance. Calculate the square root of the variance: \(SD(X) = \sqrt{Var(X)} = \sqrt{\frac{3}{2}}\) So, the standard deviation of the probability distribution is \(\sqrt{\frac{3}{2}}\).

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Key Concepts

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

Mean (Expected Value)
The mean, often called the expected value in probability distributions, is a central concept that gives us the average or central tendency of the distribution. Imagine it as the balance point of the probability distribution.
To find the mean of a probability distribution, we use the formula:
  • \( E(X) = \sum xP(X=x) \)
In this formula, you multiply each possible value of \( x \) by its probability \( P(X=x) \), and then sum all those products.
This formula gives us a weighted average, where weights are the probabilities of each possible outcome.
For the exercise example, the calculated mean was 0, meaning the distribution is perfectly balanced around this point.
Variance
Variance measures how much the values of the distribution differ from the mean. It indicates the spread or dispersion of the distribution.
To calculate variance, we use the formula:
  • \( Var(X) = E(X^2) - [E(X)]^2 \)
This involves two main steps:
First, calculate \( E(X^2) \), which is the expected value of the squared variable. It's found by squaring each value of \( x \), multiplying it by the probability \( P(X=x) \), and summing these results.
Second, subtract the square of the mean from \( E(X^2) \).
In our exercise, the variance was calculated to be \( \frac{3}{2} \), indicating how much the values vary around the mean of 0.
Standard Deviation
Standard deviation gives us a measure of the average distance of each data point from the mean.
It is simply the square root of the variance.
This makes it easier to interpret, as it brings the measure back to the same units as the original data.
The formula is straightforward:
  • \( SD(X) = \sqrt{Var(X)} \)
In the exercise, the standard deviation was found to be \( \sqrt{\frac{3}{2}} \).
This value tells us how spread out the values in the distribution are around the mean.
A lower standard deviation indicates that the data points are closer to the mean, while a higher value suggests more spread.

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Most popular questions from this chapter

The medical records of infants delivered at Kaiser Memorial Hospital show that the infants' lengths at birth (in inches) are normally distributed with a mean of 20 and a standard deviation of \(2.6\). Find the probability that an infant selected at random from among those delivered at the hospital measures a. More than 22 in. b. Less than 18 in. c. Between 19 and 21 in.

In an examination given to a class of 20 students, the following test scores were obtained: \(\begin{array}{lllllllllr}40 & 45 & 50 & 50 & 55 & 60 & 60 & 75 & 75 & 80 \\\ 80 & 85 & 85 & 85 & 85 & 90 & 90 & 95 & 95 & 100\end{array}\) a. Find the mean (or average) score, the mode, and the median score. b. Which of these three measures of central tendency do you think is the least representative of the set of scores?

The relative humidity, in percent, in the morning for the months of January through December in Boston follows: \(\begin{array}{llllll}68 & 67 & 69 & 69 & 71 & 73 \\ 74 & 76 & 79 & 77 & 74 & 70\end{array}\) Find the average and the median of these humidity readings.

The IQs of students at Wilson Elementary School were measured recently and found to be normally distributed with a mean of 100 and a standard deviation of 15 . What is the probability that a student selected at random will have an \(\mathrm{IQ}\) a. Of 140 or higher? b. Of 120 or higher? c. Between 100 and 120 ? d. Of 90 or less?

Use the appropriate normal distributions to approximate the resulting binomial distributions. Because of late cancellations, Neptune Lines, an operator of cruise ships, has a policy of accepting more reservations than there are accommodations available. From experience, \(8 \%\) of the bookings for the 90-day around-the- world cruise on the S.S. Drion, which has accommodations for 2000 passengers, are subsequently canceled. If the management of Neptune Lines has decided, for public relations reasons, that \(99 \%\) of all booked passengers will obtain accommodation on the ship, determine the largest number of reservations that should be taken for this cruise on the S.S. Drion.
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