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Chapter 5: Problem 29

Let \(x\) be the number of heads obtained in two tosses of a coin. The following table lists the probability distribution of \(x\). $$ \begin{array}{l|lll} \hline x & 0 & 1 & 2 \\ \hline P(x) & .25 & .50 & .25 \\ \hline \end{array} $$ Calculate the mean and standard deviation of \(x\). Give a brief interpretation of the value of the mean.

Short Answer

Expert verified
The mean or expected value of the distribution is .75, implying that we would expect, on average, to get heads approximately .75 times if we repeated the experiment. The standard deviation is also .75, indicating a 75% variation around the mean.

Step by step solution

01

Calculate the Mean

The mean, often represented as \( \mu \), of a random variable \( x \) is given by the sum of the product of each outcome and its probability. Using the given distribution, the calculation is as follows: \( \mu = 0(.25) + 1(.50) + 2(.25) = .75 \)
02

Calculate the standard deviation

The standard deviation, often represented by \( \sigma \), quantifies the amount of variation in the distribution. It is calculated as follows: \( \sigma = \sqrt{\Sigma(P(x).(x-\mu)^2)} \). Plugging the given values and the calculated mean, we get: \( \sigma = \sqrt{(0 - .75)^2 * .25 + (1 - .75)^2 * .50 + (2 - .75)^2 * .25} = \sqrt{.5625} \) which simplifies to approximately \( \sigma = .75 \)
03

Interpret the Mean

The mean or expected value is the average outcome we would expect if we repeat the experiment many times. So in this case, if we toss the coin two times, on average we would expect to get heads approximately .75 times. This is not possible for a single experiment but it makes sense over a large number of experiments. In general, a higher mean indicates that the outcome of getting heads is more likely.

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Key Concepts

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

Mean Calculation
The mean, also known as the expected value, is a fundamental concept in probability distribution. It gives us an idea of the average outcome in a repeated experiment. For a random variable like the number of heads in a coin toss, finding the mean involves multiplying each outcome by its probability and summing up these products. This represents what you might expect on average after many repetitions.
To calculate it:
  • Take each possible outcome, like obtaining 0, 1, or 2 heads.
  • Multiply each outcome by its probability. Here, you have 0 heads with a probability of 0.25, 1 head with a probability of 0.50, and 2 heads with a probability of 0.25.
  • Add these products together: \( \mu = 0 \times 0.25 + 1 \times 0.50 + 2 \times 0.25 = 0.75 \)
This mean of 0.75 does not imply that you will see exactly 0.75 heads in each trial but represents the long-term average of heads per set of two coin tosses.
Standard Deviation
Standard deviation is a measure that is used to quantify the amount of variation or dispersion in a set of values. In the context of probability distribution, it tells us how much the outcomes deviate from the mean on average.
To calculate the standard deviation for our coin toss problem:
  • Determine the mean, which is 0.75 as calculated earlier.
  • Subtract the mean from each outcome to find the deviation of each outcome.
  • Square each deviation to avoid negative numbers and multiply it by the probability of the corresponding outcome.
  • Sum these values and take the square root: \( \sigma = \sqrt{(0 - 0.75)^2 \times 0.25 + (1 - 0.75)^2 \times 0.50 + (2 - 0.75)^2 \times 0.25} = \sqrt{0.5625} \approx 0.75 \)
This tells us that the outcomes, on average, deviate from the mean by about 0.75 heads in this case.
Interpretation of Mean
Interpreting the mean, especially in probability, can sometimes be tricky. The calculated mean of 0.75 heads for the two tosses of a coin gives you an insight into the expected outcome over a large number of tosses. Even though you cannot achieve 0.75 heads in a single set of two tosses, this figure becomes meaningful as the number of sets increases.
Think of the mean as the long-term average:
  • If you conduct many sets of two coin tosses, the average number of heads will approach 0.75 per set.
  • This does not predict individual results but guides your understanding of trends over time.
Higher mean values would signal that getting more heads is likely more favorable, based on the distribution of probabilities across different outcomes.

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Most popular questions from this chapter

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