Problem 20 - Calculus for Biology and Medicine

Question

Suppose \(S_{n}\) is binomially distributed with parameters \(n=\) 150 and \(p=0.4\). Use the central limit theorem to find an approximation for \(P\left(S_{n}=60\right)\) (a) without the histogram correction and (b) with the histogram correction. (c) Use a graphing calculator to compute the exact probabilities and compare your answers with those in (a) and (b)

Step-by-Step Solution

Answer

(a) Definitionally zero; (b) 0.0668; (c) Exact: 0.0577. Histogram correction gives closer approximation.

Step 1: Understand the Problem

We are given a binomial distribution with parameters \( n = 150 \) and \( p = 0.4 \). We need to approximate \( P(S_n = 60) \) using the Central Limit Theorem (CLT) in two ways: (a) without histogram correction, and (b) with histogram correction. Finally, we'll compare these approximations with the exact probability using a graphing calculator.

Step 2: Calculate Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) is given by \( \mu = np = 150 \times 0.4 = 60 \). The standard deviation \( \sigma \) is given by \( \sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.4 \times 0.6} \approx 6 \).

Step 3: Approximation without Histogram Correction

By the Central Limit Theorem, the distribution of \( S_n \) can be approximated by a normal distribution with mean \( \mu = 60 \) and standard deviation \( \sigma = 6 \). However, finding \( P(S_n = 60) \) in a continuous distribution directly isn't possible, so traditionally this would be the area over an infinitesimally small interval, which leads to very small results.

Step 4: Approximation with Histogram Correction

With histogram correction (also known as continuity correction), we adjust our discrete \( x = 60 \) in the binomial to a continuous interval for the normal approximation: \( P(59.5 < X < 60.5) \). We calculate:\[ P(59.5 < X < 60.5) = P\left( \frac{59.5 - 60}{6} < Z < \frac{60.5 - 60}{6} \right) \]\[ = P(-0.0833 < Z < 0.0833) \]Using a standard normal distribution table, \( P(-0.0833 < Z < 0.0833) \approx 0.0668 \).

Step 5: Compute Exact Probability Using a Graphing Calculator

Using a graphing calculator or statistical software, calculate the exact probability \( P(S_n = 60) \). The calculator directly computes the probability for a binomial distribution, which is approximately 0.0577.

Step 6: Compare the Approximations

The approximation without histogram correction resulted in a more negligible probability. With histogram correction, the approximation was 0.0668, which is closer to the exact probability from the calculator, 0.0577. Thus, the histogram correction provides a better approximation.

Key Concepts

Binomial Distribution

The Binomial Distribution is a discrete probability distribution. It describes the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. For example, if you're flipping a coin 150 times and the probability of getting heads (success) each time is 0.4, this scenario can be modeled using a binomial distribution where \( n = 150 \) and \( p = 0.4 \).

• **n** – the number of trials
• **p** – the probability of success on each trial
• **S_n** – the random variable representing the number of successes

The formula for the probability of exactly \( k \) successes (like 60 heads) is given by:\[P(S_n = k) = \binom{n}{k} p^k (1-p)^{n-k}\]However, calculating this for large \( n \) can be computationally intensive, which is why techniques like the Central Limit Theorem are often used for approximation.

Normal Approximation

The Normal Approximation is a method used to approximate the behavior of binomial distributions. According to the Central Limit Theorem, if \( n \) is large enough, the binomial distribution can be approximated using a normal distribution with mean \( \mu = np \) and standard deviation \( \sigma = \sqrt{np(1-p)} \).
In the given exercise, this translates to approximating \( S_n \) with a normal distribution having:

• Mean: \( \mu = 60 \)
• Standard Deviation: \( \sigma = 6 \)

To find \( P(S_n = 60) \), without histogram correction, you're looking for a very narrow area under the curve, almost like a vertical line, which results in an essentially negligible probability. This is why histogram correction is preferred when making such approximations.

Histogram Correction

Histogram Correction, also known as continuity correction, is a technique to improve the approximation of a discrete probability by a continuous one. This correction essentially aligns the discrete nature of binomial distributions with the continuous nature of normal distributions.
In continuous probability distributions like the normal distribution, you need to consider an interval instead of an exact point when converting from a discrete distribution. For the given \( S_n = 60 \), you use the interval \( P(59.5 < X < 60.5) \) instead of \( P(X = 60) \).
This adjustment captures the area under the normal curve over an interval of half a unit above and below 60, allowing us to approximate:\[P\left( \frac{59.5 - 60}{6} < Z < \frac{60.5 - 60}{6} \right) = P(-0.0833 < Z < 0.0833)\]Using standard normal tables or software, you can find that this probability is roughly 0.0668. This result is usually considered more accurate than the uncorrected version.

Probability Approximation

Probability Approximation helps to compute probabilities in binomial distributions more efficiently, especially for larger \( n \). Without it, calculating probabilities for specific outcomes could be cumbersome and time-consuming.

• **Exact Methods**: Direct computation using binomial formula can be slow.
• **Approximate Methods**: Techniques like the Normal Approximation simplify calculations.
• **Histogram Correction**: This further refines the approximation.

Ultimately, using a graphing calculator or statistical software is often a convenient way to compute exact probabilities. In our exercise, the exact probability for \( P(S_n = 60) \) calculated as 0.0577 using these tools provides a benchmark.
The comparison emphasized that histogram correction (yielding 0.0668) was a closer approximation to the exact probability than the simple normal approximation without correction.