Problem 18
Question
Toss a fair \(\operatorname{coin} 300\) times. (a) Use the central limit theorem and the histogram correction to find an approximation for the probability that the number of heads is between 140 and 160 . (b) Use Chebyshev's inequality to find an estimate for the event in (a), and compare your estimate with that in (a).
Step-by-Step Solution
Verified Answer
The probability of having between 140 to 160 heads is approximately 0.7740 via the normal approximation; Chebyshev's inequality estimates it as at least 0.248.
1Step 1: Define the variables and problem
For a fair coin tossed 300 times, we expect heads in each trial with probability \( p = 0.5 \). Therefore, we are dealing with a binomial distribution \( X \sim \text{Bin}(300, 0.5) \). We need to find the probability that the number of heads is between 140 and 160, i.e., \( P(140 \leq X \leq 160) \).
2Step 2: Apply Central Limit Theorem (CLT)
According to the central limit theorem, for large \( n \), the binomial distribution can be approximated by a normal distribution \( N(\mu, \sigma^2) \) where \( \mu = np \) and \( \sigma = \sqrt{np(1-p)} \). Here, \( \mu = 300 \times 0.5 = 150 \) and \( \sigma = \sqrt{300 \times 0.5 \times (1-0.5)} = \sqrt{75} \approx 8.66 \).
3Step 3: Apply Histogram Correction
When using the normal approximation for a discrete distribution, apply a continuity correction. Thus, find \( P(139.5 < X < 160.5) \). Convert it to the standard normal variable \( Z \) with \( Z = \frac{X - \mu}{\sigma} \). Calculate \( P \left( \frac{139.5 - 150}{8.66} < Z < \frac{160.5 - 150}{8.66} \right) \).
4Step 4: Calculate the Normal Approximation
The standard normal values are: \( \frac{139.5 - 150}{8.66} \approx -1.21 \) and \( \frac{160.5 - 150}{8.66} \approx 1.21 \). Use standard normal distribution tables to find probabilities. \( P(Z < 1.21) \approx 0.8871 \) and \( P(Z < -1.21) \approx 0.1131 \). The probability \( P(-1.21 < Z < 1.21) = 0.8871 - 0.1131 = 0.7740 \).
5Step 5: Apply Chebyshev's Inequality
Chebyshev’s inequality states \( P(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \). Here, \( \mu = 150 \), \( \sigma = 8.66 \), and we want \( 140 \leq X \leq 160 \); hence, \( k = \frac{10}{8.66} \approx 1.154 \). Calculate \( P(|X - 150| < 10) \geq 1 - \frac{1}{1.154^2} \approx 0.248 \).
6Step 6: Compare Results
The probability from the normal approximation is approximately 0.7740, whereas Chebyshev's inequality provides a lower bound probability of at least 0.248. The normal approximation is more precise and useful for this specific problem.
Key Concepts
Binomial DistributionNormal ApproximationChebyshev's Inequality
Binomial Distribution
The binomial distribution is a discrete probability distribution. It models the number of successes in a fixed number of independent trials, with each trial having two possible outcomes. This is similar to tossing a coin multiple times and counting how many times it lands heads up.
Key characteristics of a binomial distribution include:
Key characteristics of a binomial distribution include:
- Number of trials, denoted as \( n \.\)
- Probability of success on each trial, \( p \,\)
- The formula for the probability of exactly \( k \,\) successes is given by the binomial formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \.\]
Normal Approximation
The normal approximation is a technique used to approximate a binomial distribution with a normal distribution. This is especially useful when dealing with a large number of trials because the calculations using binomial distribution directly become cumbersome.
The Central Limit Theorem (CLT) provides a foundation for this approximation, stating that as the number of trials increase, the binomial distribution will tend to the shape of a normal distribution. To apply this:
The Central Limit Theorem (CLT) provides a foundation for this approximation, stating that as the number of trials increase, the binomial distribution will tend to the shape of a normal distribution. To apply this:
- Calculate the mean \( \mu = np \,\) and the variance \( \sigma^2 = np(1-p) \.\)
- Use these to form a normal distribution \( N(\mu, \sigma^2) \.\)
- Apply the continuity correction when dealing with discrete data by adjusting boundaries by 0.5.
Chebyshev's Inequality
Chebyshev's inequality is a statistical tool that provides a bound on the probability that a random variable deviates from its mean. It does not require the data to follow any specific distribution, making it widely applicable.
The inequality states that for a random variable \( X \) with mean \( \mu \,\) and standard deviation \( \sigma \,\), the probability that \( X \) falls within \( k \sigma \) of the mean is at least \( 1 - \frac{1}{k^2} \.\)
The inequality states that for a random variable \( X \) with mean \( \mu \,\) and standard deviation \( \sigma \,\), the probability that \( X \) falls within \( k \sigma \) of the mean is at least \( 1 - \frac{1}{k^2} \.\)
- Sample application: \( P(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \,\)
- It gives a minimum probability that a certain event will be within a specified range.
Other exercises in this chapter
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Assume that \(P(A)=0.4, P(B)=0.4\), and \(P(A \cup B)=0.7\). Find \(P(A \cap B)\) and \(P\left(A^{c} \cap B^{c}\right)\).
View solution