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Chapter 2: Problem 67

Suppose that \(X\) is a random variable with mean 10 and variance 15 . What can we say about \(P[5

Short Answer

Expert verified
Using Chebyshev's inequality, we find that the probability of the random variable X being between 5 and 15 is at least \(\frac{2}{5}\) or 0.4. This is a lower bound, and the actual probability could be higher, depending on the specific distribution of the random variable X.

Step by step solution

01

Understand Chebyshev's inequality

Chebyshev's inequality states that for any random variable X with mean 渭 and variance 蟽虏, the probability that X is within k standard deviations (where k > 0) of the mean is greater or equal to the value: \[P(|X - \mu| \le k\sigma) \ge 1 - \frac{1}{k^2}\]
02

Calculate the standard deviation

To apply Chebyshev's inequality, we first need to calculate the standard deviation (蟽) of the random variable X. The standard deviation is the square root of the variance: \[\sigma = \sqrt{\sigma^2} = \sqrt{15}\]
03

Find k that satisfies the given interval

We want to find the probability that X lies between 5 and 15, i.e., \(P(5 < X < 15)\). Let's express this in terms of the mean and standard deviation: \(10 - k\sqrt{15} < X < 10 + k\sqrt{15}\) Now we need to find the value of k that fits this interval. Since both sides are equidistant from the mean, we can set up the equation: \(10 - 5 = k\sqrt{15}\) \(k\sqrt{15} = 5\) Now solve for k: \(k = \frac{5}{\sqrt{15}}\)
04

Apply Chebyshev's inequality

Now we plug the value of k into Chebyshev's inequality: \[P(5 < X < 15) \ge 1 - \frac{1}{k^2}\] Plugging in the value of k: \[P(5 < X < 15) \ge 1 - \frac{1}{\left(\frac{5}{\sqrt{15}}\right)^2}\] Solve: \[P(5 < X < 15) \ge 1 - \frac{1}{\left(\frac{25}{15}\right)}\] \[P(5 < X < 15) \ge 1 - \frac{15}{25}\] \[P(5 < X < 15) \ge \frac{10}{25}\] \[\boxed{P(5 < X < 15) \ge \frac{2}{5}}\] So we can say that the probability of X being between 5 and 15 is at least 2/5 or 0.4. It's important to remember that this is a lower bound, and the actual probability could be higher, depending on the specific distribution of the random variable X.

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Key Concepts

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

Variance
Variance is a measure of how much a set of numbers varies from its average (mean). For a random variable like \(X\), variance (\( \sigma^2 \)) helps us understand the spread or dispersion of data around the mean.

To calculate variance, follow these steps:
  • Find the mean (\( \mu \)) of the dataset.
  • Subtract the mean from each data point and square the result.
  • Average these squared differences.
This gives you the variance.

In our example, the variance is 15, meaning the data is moderately spread out from the mean of 10. A higher variance would signify a larger spread.
Standard Deviation
Standard deviation is the square root of variance. It provides a direct measure of how much individual data points in a dataset typically deviate from the mean.

To find standard deviation:
  • Take the square root of the variance.
For our problem, if the variance (\( \sigma^2 \)) is 15, then the standard deviation (\( \sigma \)) is \( \sqrt{15} \approx 3.87 \).

Standard deviation is easier to relate to than variance because it is in the same units as the original data, making it more intuitive to understand and interpret. It tells us how much we can expect individual data points to differ from the mean.
Probability Bounds
Probability bounds refer to the range within which a random variable is likely to fall. In this context, Chebyshev's Inequality is used to determine these bounds.

Chebyshev's Inequality provides a way to estimate the probability that a random variable is a certain number of standard deviations from the mean:
  • The inequality is \( P(|X - \mu| \le k\sigma) \ge 1 - \frac{1}{k^2} \).
  • The inequality holds for any random variable, regardless of its distribution, making it very flexible.
Using Chebyshev's Inequality, you can estimate bounds with just the mean, variance, and the desired interval said in terms of standard deviations (like in our solution where it was used to estimate the probability of \(X\) falling between 5 and 15).
Random Variable
A random variable is a function that assigns numerical values to the outcomes of a random phenomenon. In simple terms, it's a variable whose outcomes depend on the result of a random event.

There are different types of random variables:
  • Discrete random variables, which have specific values like 1, 2, 3, etc.
  • Continuous random variables, which can take any value within a range.
In our exercise, \(X\) is a random variable with a mean of 10, representing its expected value over many trials.

The random variable's properties are essential for solving problems involving probability distributions and probability bounds. Understanding these concepts helps us make predictions about the potential outcomes of \(X\).

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Most popular questions from this chapter

An urn contains \(n+m\) balls, of which \(n\) are red and \(m\) are black. They are withdrawn from the urn, one at a time and without replacement. Let \(X\) be the number of red balls removed before the first black ball is chosen. We are interested in determining \(E[X]\). To obtain this quantity, number the red balls from 1 to \(n\). Now define the random variables \(X_{i}, i=1, \ldots, n\), by \(X_{i}=\left\\{\begin{array}{ll}1, & \text { if red ball } i \text { is taken before any black ball is chosen } \\ 0, & \text { otherwise }\end{array}\right.\) (a) Express \(X\) in terms of the \(X_{1}\). (b) Find \(E[X]\).

Let \(X\) and \(Y\) each take on either the value 1 or \(-1\). Let $$ \begin{array}{r} p(1,1)=P[X=1, Y=1] \\ p(1,-1)=P[X=1, Y=-1] \\ p(-1,1)=P \mid X=-1, Y=1] \\ p(-1,-1)=P[X=-1, Y=-1] \end{array} $$ Suppose that \(E[X]=E[Y]=0 .\) Show that (a) \(p(1,1)=p(-1,-1)\) (b) \(p(1,-1)=p(-1,1)\) Let \(p=2 p(1,1) .\) Find (c) \(\operatorname{Var}(X)\) (d) \(\operatorname{Var}(Y)\) (e) \(\operatorname{Cov}(X, Y)\)

Suppose \(X\) has a binomial distribution with parameters 6 and \(\frac{1}{2}\). Show that \(X=3\) is the most likely outcome.

An urn contains five red, three orange, and two bluc balls. Two balls are randomly selected. What is the sample space of this experiment? Let \(X\) represent the number of orange balls selected. What are the possible values of \(X\) ? Calculate \(P[X=0\\}\).

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their policy is to sell 52 tickets for a flight that can only hold 50 passengers. What is the probability that there will be a seat available for cvery passenger that shows up?
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