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Chapter 6: Problem 22

Random numbers Let \(Y\) be a number between 0 and 1 produced by a random number generator. Assuming that the random variable \(Y\) has a uniform distribution, find the following probabilities: \(\begin{array}{ll}\text { (a) } & P(Y \leq 0.4)\end{array}\) (b) \(P(Y<0.4)\) \(\begin{array}{ll}\text { (c) } & P(0.1

Short Answer

Expert verified
(a) 0.4, (b) 0.4, (c) 0.16

Step by step solution

01

Understand the Uniform Distribution

Since the random variable \(Y\) is uniformly distributed over \([0,1]\), its probability density function is constant over the interval \([0, 1]\). This makes the distribution easy to handle because the probability of \(Y\) falling into any sub-interval \([a,b]\) is simply \(b-a\).
02

Calculate \(P(Y \leq 0.4)\)

In a uniform distribution over \([0,1]\), the probability \(P(Y \leq 0.4)\) is equal to the length of the interval from \(0\) to \(0.4\). Therefore, \(P(Y \leq 0.4) = 0.4 - 0 = 0.4\).
03

Calculate \(P(Y < 0.4)\)

For continuous random variables, \(P(Y \leq 0.4)\) and \(P(Y < 0.4)\) are equal because the probability of \(Y\) being exactly \(0.4\) is zero. Hence, \(P(Y < 0.4) = 0.4\).
04

Calculate \(P(0.1 < Y \leq 0.15 \, \text{or} \, 0.77 \leq Y < 0.88)\)

First, find the probability of \(Y\) in the interval \(0.1 < Y \leq 0.15\), which is \(0.15 - 0.1 = 0.05\). Next, find the probability of \(Y\) in the interval \(0.77 \leq Y < 0.88\), which is \(0.88 - 0.77 = 0.11\). Since these two intervals do not overlap, the total probability is the sum of the two separate probabilities, \(0.05 + 0.11 = 0.16\).

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Key Concepts

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

Random Variable
A random variable is a crucial concept in probability and statistics. It refers to a numerical outcome that results from a random event.
In simpler terms, you can think of it as a variable whose value is determined by the outcome of a random process or experiment.

For example:
  • When you toss a coin, if you define "Heads" as 1 and "Tails" as 0, the result of the toss is a random variable.
  • In our exercise, the random variable is represented by \(Y\), which can take any value between 0 and 1, inclusive.
These values are not chosen methodically but are generated randomly, following the distribution rules specified.
Probability Calculation
Probability calculation is a way to quantify how likely an event will occur. In the context of a uniform distribution, the calculations simplify considerably. If you know the interval to which your random variable is confined, calculating probabilities becomes straightforward.

For instance:
  • If a number \(Y\) is uniformly distributed between 0 and 1, the probability \(P(Y \leq 0.4)\) is simply the width of the interval from 0 to 0.4, which equals 0.4.
  • Similarly, the probability \(P(Y < 0.4)\) is also 0.4 because with continuous random variables, the calculation \(P(Y \leq 0.4)\) and \(P(Y < 0.4)\) yield identical results.
Thus, each outcome has an equal chance of occurring and the probability of an interval matches exactly its length in the distribution's range.
Interval Probability
Understanding interval probability is critical when working with continuous distributions.
Instead of looking at individual outcomes, we look at the probability of the random variable falling within a certain range or interval.
For example:
  • In the problem above, we had the intervals \(0.1 < Y \leq 0.15\) and \(0.77 \leq Y < 0.88\). To find the total probability in these non-overlapping intervals, we calculate each separately and then sum them up.
  • The probability for each interval equals the interval's length, so for \(0.1 < Y \leq 0.15\), it is \(0.05\), and for \(0.77 \leq Y < 0.88\), it is \(0.11\).
Adding these probabilities gives us the total interval probability of \(0.16\). Intervals, thus, provide a powerful way to explore probabilities over continuous ranges.
Continuous Distribution
In probability theory, a continuous distribution describes the probabilities of the possible values of a continuous random variable.
For the uniform distribution, every outcome within the interval is equally likely.

Some key characteristics are:
  • Its probability density function is constant over its support, which in this case is the interval \([0, 1]\). This means any small segment of the interval has the same likelihood.
  • In a continuous distribution, the probability of the random variable taking an exact value is zero because there are infinitely many potential values.
In practical applications, such as random number generation between 0 and 1, continuous distributions like the uniform distribution are used to represent data that can take on any value within a defined range. This makes them ideal for simulations and statistical modeling.

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Most popular questions from this chapter

His and her earnings Researchers randomly select a married couple in which both spouses are employed. Let \(X\) be the income of the husband and \(Y\) be the income of the wife. Suppose that you know the means \(\mu_{X}\) and \(\mu_{Y}\) and the variances \(\sigma_{X}^{2}\) and \(\sigma_{Y}^{2}\) of both variables. (a) Is it reasonable to take the mean of the total income \(X+Y\) to be \(\mu_{X}+\mu_{Y} ?\) Explain your answer. (b) Is it reasonable to take the variance of the total income to be \(\sigma_{X}^{2}+\sigma_{Y}^{2}\) ? Explain your answer.

Marti decides to keep placing a \(\$ 1\) bet on number 15 in consecutive spins of a roulette wheel until she wins. On any spin, there's a 1 -in- 38 chance that the ball will land in the 15 slot. (a) How many spins do you expect it to take until Marti wins? Justify your answer. (b) Would you be surprised if Marti won in 3 or fewer spins? Compute an appropriate probability to support your answer.

Rhubarb Suppose you purchase a bundle of 10 bareroot rhubarb plants. The sales clerk tells you that \(5 \%\) of these plants will die before producing any rhubarb. Assume that the bundle is a random sample of plants and that the sales clerk's statement is accurate. Let \(Y=\) the number of plants that die before producing any rhubarb. Use the binomial probability formula to find \(P(Y=1) .\) Interpret this result in context.

determine whether the given random variable has a binomial distribution. Justify your answer. Sowing seeds Seed Depot advertises that its new flower seeds have an \(85 \%\) chance of germinating (growing). Suppose that the company's claim is true. Judy gets a packet with 20 randomly selected new flower seeds from Seed Depot and plants them in her garden. Let \(X=\) the number of seeds that germinate.

Men's heights A report of the National Center for Health Statistics says that the height of a 20 -year-old man chosen at random is a random variable \(H\) with mean 5.8 feet and standard deviation 0.24 feet. Find the mean and standard deviation of the height \(J\) of a randomly selected 20 -year-old man in inches. There are 12 inches in a foot.
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