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Chapter 8: Problem 20

Let the random variable \(X\) denote the number of girls in a five-child family. If the probability of a female birth is .5, a. Find the probability of \(0,1,2,3,4\), and 5 girls in a fivechild family. b. Construct the binomial distribution and draw the histogram associated with this experiment. c. Compute the mean and the standard deviation of the random variable \(X\).

Short Answer

Expert verified
a. The probabilities are: - P(0 girls) = \( \binom{5}{0}(0.5)^0(1-0.5)^{5-0} \) = 0.03125 - P(1 girl) = \( \binom{5}{1}(0.5)^1(1-0.5)^{5-1} \) = 0.15625 - P(2 girls) = \( \binom{5}{2}(0.5)^2(1-0.5)^{5-2} \) = 0.3125 - P(3 girls) = \( \binom{5}{3}(0.5)^3(1-0.5)^{5-3} \) = 0.3125 - P(4 girls) = \( \binom{5}{4}(0.5)^4(1-0.5)^{5-4} \) = 0.15625 - P(5 girls) = \( \binom{5}{5}(0.5)^5(1-0.5)^{5-5} \) = 0.03125 b. Histogram: X-axis represents the number of girls (0-5), and Y-axis represents the probabilities. Draw bars with heights equal to their respective probabilities. c. Mean (碌) = 5 * 0.5 = 2.5; Standard deviation (蟽) = \( \sqrt{5 * 0.5 * (1 - 0.5)} \) = 1.118.

Step by step solution

01

a. Calculate the probabilities.

To find the probabilities of having 0, 1, 2, 3, 4, and 5 girls in a five-child family, we'll use the binomial probability formula: \[ P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \] where n is the number of trials (children in this case), k is the number of successes (girls), p is the probability of success (female birth), and \( \binom{n}{k} \) is the binomial coefficient given as \( \f rac{n!}{k!(n-k)!} \). We'll plug in the given values, where \( n = 5 \) and \( p = 0.5 \), and calculate the probabilities for \( k = 0, 1, 2, 3, 4, \) and 5.
02

b. Construct the binomial distribution and draw the histogram.

Now that we have the probabilities, we can create a binomial distribution table: | X (number of girls) | P(X = k) | |----------------------|------------------------------| | 0 | (calculated probability) | | 1 | (calculated probability) | | 2 | (calculated probability) | | 3 | (calculated probability) | | 4 | (calculated probability) | | 5 | (calculated probability) | To draw the histogram, put X values (number of girls) on the horizontal axis and their corresponding probabilities (P(X = k)) on the vertical axis. For each X value, draw a bar with a height equal to its respective probability.
03

c. Compute the mean and the standard deviation.

To find the mean and standard deviation of the given binomial distribution, we'll apply the formulas: Mean (碌) = n * p Standard deviation (蟽) = \( \sqrt{n * p * (1 - p)} \) Plug in the given values \( n = 5 \) and \( p = 0.5 \) to calculate the mean and standard deviation.

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Key Concepts

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

Probability Theory
Probability theory helps us understand the likelihood of occurrences in random experiments. It uses a range of methods and formulas to predict the outcomes of uncertain events. One common application is determining the probabilities of various outcomes of a random variable.

In our example, we're given a scenario of a five-child family where the probability of having a girl (success) is 0.5 for each child. Using probability theory, we can calculate the likelihood of having 0 to 5 girls in the family. This is achieved with the binomial probability formula:
  • The formula is: \( P(X = k) = \binom{n}{k}p^k(1-p)^{n-k} \)
  • \( n \) is the number of trials (children).
  • \( k \) is the number of successes (the number of girls).
  • \( p \) is the probability of a success (birth of a girl).
By plugging in the values, such as \( n = 5 \) and \( p = 0.5 \), we find the probabilities of having different numbers of girls in the family.
Random Variables
A random variable is a numerical outcome of a random phenomenon. It maps outcomes of a random process to numbers. In the binomial distribution context, the random variable often represents the number of successes in trials.

In our problem, the random variable \( X \) is defined as the number of girls in a five-child family. The possible values \( X \) can take range from 0 to 5, depending on how many girls there are in the family. These values are finite and discrete.

Understanding random variables in probability theory helps us make predictions about future events based on past occurrences. It turns real-world phenomena into a mathematical model, allowing us to perform computations and draw useful conclusions about probability distributions.
Mean and Standard Deviation
The mean and standard deviation are two vital concepts in statistics used to describe a probability distribution.

Mean (Expected Value)

The mean of a random variable in a probability distribution tells us the average outcome. It's calculated as the sum of all possible values of the variable, each multiplied by its probability.

For a binomial distribution, the mean \( \mu \) is given by the product of the number of trials \( n \) and the probability of success \( p \):
  • \( \mu = n \cdot p \)
In our scenario, with \( n = 5 \) and \( p = 0.5 \), the mean number of girls is 2.5.

Standard Deviation

The standard deviation measures the spread or dispersion of the random variable around the mean. A higher value signifies more variability in the outcomes. For a binomial distribution, the standard deviation \( \sigma \) is given by:
  • \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} \)
Here, this results in approximately 1.118, indicating how much the number of girls might vary from the average.

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Most popular questions from this chapter

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