Chapter 27: Problem 52
Let a random variable X have a binomial distribution with mean 8 and variance 4. If \(\mathrm{P}\left(\mathrm{X} d^{\prime \prime} 2\right)=\frac{k}{2^{16}}\), then \(\mathrm{k}\) is equal to: \([\) April 12, \(2019(\mathrm{I})]\) (a) 17 (b) 121 (c) 1 (d) 137
Short Answer
Expert verified
The value of \(k\) is 137.
Step by step solution
01
Understand the given mean and variance
The mean of a binomial distribution is calculated by \(np\), and the variance is given by \(np(1-p)\), where \(n\) is the number of trials and \(p\) is the probability of success in each trial. From the problem, we know \(np = 8\) and \(np(1-p) = 4\).
02
Solve for p
First, express \(p\) as \(p = \frac{8}{n}\) using the mean \(np = 8\). Substitute \(p\) into the variance equation: \(\frac{8}{n}(1 - \frac{8}{n}) = \frac{4}{n}\). Simplify to get the quadratic equation \(n^2 - 8n = 16\).
03
Solve the quadratic equation
Solve \(n^2 - 8n - 16 = 0\). Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -8, c = -16\), find the solutions for \(n\).
04
Determine integer values for n and p
Solving the quadratic equation gives us \(n = 10\) as the valid positive integer solution (discard any negative or non-integer solutions). Substituting back, \(p = \frac{8}{10} = 0.8\).
05
Substitute n and p into binomial PMF
The binomial probability mass function (PMF) is given by \(\mathrm{P}(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\). Substitute \(n = 10\) and \(p = 0.8\), and \(x = 2\) into the PMF formula.
06
Calculate the probability P(X=2)
Calculate \(\mathrm{P}(X = 2) = \binom{10}{2} (0.8)^2 (0.2)^8\). This equals to \(\frac{k}{2^{16}}\), by evaluating the binomial coefficient and powers.
07
Solve for k
Use \(\binom{10}{2} = 45\), calculate \(\mathrm{P}(X = 2) = 45 \times (0.8)^2 \times (0.2)^8\). This can be rearranged to solve \(\frac{k}{2^{16}} = 45 \times 0.64 \times (0.2)^8\). Calculate and simplify to find \(k = 137\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
Quadratic equations are mathematical expressions that take the form \( ax^2 + bx + c = 0 \). These equations involve an unknown variable \( x \) raised to the second power, leading to a parabola when graphed. They often appear in problems where relationships are not linear, such as our binomial distribution exercise.
To solve a quadratic equation, you can use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the values of \( x \) by considering the coefficients \( a \), \( b \), and \( c \) from the equation. First, determine the discriminant, \( b^2 - 4ac \), which indicates the nature of the roots:
To solve a quadratic equation, you can use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the values of \( x \) by considering the coefficients \( a \), \( b \), and \( c \) from the equation. First, determine the discriminant, \( b^2 - 4ac \), which indicates the nature of the roots:
- Positive discriminant means two real and different roots
- Zero discriminant means one real and repeated root
- Negative discriminant means complex roots
Probability Mass Function
The probability mass function (PMF) expresses the probability that a discrete random variable is exactly equal to a certain value. For a binomial distribution, which models the number of successes in a series of independent experiments, the PMF is essential in calculating probabilities.
The PMF for a binomial distribution is given by:\[P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\]Here:
The PMF for a binomial distribution is given by:\[P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\]Here:
- \( \binom{n}{x} \) is a binomial coefficient that counts the number of ways to choose \( x \) successes in \( n \) trials
- \( p \) is the probability of success on each trial
- \( 1-p \) is the probability of failure
- \( x \) is the number of successes
Mean and Variance of Binomial Distribution
The mean and variance are fundamental characteristics of a binomial distribution. They provide insights into the expected outcome and the variability of the distribution:
- **Mean** \( \mu \): This represents the average number of successes and is calculated by the formula \( np \).- **Variance** \( \sigma^2 \): This indicates the spread or variability in a distribution and is given by \( np(1-p) \).
For a binomial distribution with number of trials \( n \) and success probability \( p \), both mean and variance are directly linked to \( n \) and \( p \):
- **Mean** \( \mu \): This represents the average number of successes and is calculated by the formula \( np \).- **Variance** \( \sigma^2 \): This indicates the spread or variability in a distribution and is given by \( np(1-p) \).
For a binomial distribution with number of trials \( n \) and success probability \( p \), both mean and variance are directly linked to \( n \) and \( p \):
- Mean \( = np \)
- Variance \( = np(1-p) \)