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

Without correction, approximation is 0. With correction, it's 0.076. Exact is 0.060.

Step 1: Identify the Given Parameters

We are given a binomial distribution with parameters \( n = 150 \) and \( p = 0.4 \). We need to find \( P(S_n = 60) \).

Step 2: Calculate Mean and Standard Deviation

The mean \( \mu \) of the binomial distribution is given by \( \mu = n \times p = 150 \times 0.4 = 60 \). The standard deviation \( \sigma \) is \( \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{150 \times 0.4 \times 0.6} \approx 5.196 \).

Step 3: Approximate Probability without Histogram Correction

Using the central limit theorem, the probability \( P(S_n = 60) \) can be approximated by the normal distribution as \( P\left(Z = \frac{60 - 60}{5.196} \right) = P(Z = 0) \). For a standard normal distribution, \( P(Z = 0) = 0 \) because the probability at a single point for a continuous distribution is zero.

Step 4: Approximate Probability with Histogram Correction

For the histogram correction (continuity correction), we look at the probability of \( P(59.5 < S_n < 60.5) \). Calculate standardized values: \( Z_1 = \frac{59.5 - 60}{5.196} \approx -0.096 \) and \( Z_2 = \frac{60.5 - 60}{5.196} \approx 0.096 \). Find \( P(-0.096 < Z < 0.096) \) using the standard normal table. \( P(-0.096 < Z < 0.096) \approx 0.076 \).

Step 5: Compute Exact Probability using Binomial Formula

For the exact probability, use the binomial probability formula: \( P(S_n = 60) = \binom{150}{60} \times 0.4^{60} \times 0.6^{90} \). Calculate using a calculator or computer. \( P(S_n = 60) \approx 0.060 \).

Step 6: Compare Approximations with Exact Probability

From (a), with no correction, \( P(S_n = 60) \approx 0 \). With continuity correction, (b), \( P(S_n = 60) \approx 0.076 \). Exact probability from (c) is \( 0.060 \). The corrected approximation is closer to the exact value.

Key Concepts

Understanding Binomial Distribution

A binomial distribution is a discrete probability distribution. It describes the number of successes in a fixed number of trials. Each trial has two possible outcomes: success or failure. The parameters of the binomial distribution are:

• Number of trials, denoted as \( n \).
• Probability of success on each trial, denoted as \( p \).

For example, if a coin is flipped 10 times, and you want to know the likelihood of flipping heads 6 times, you can use a binomial distribution. You calculate the probability of a particular number of successes using the binomial formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( \binom{n}{k} \) is a binomial coefficient. It's calculated by \( \frac{n!}{k!(n-k)!} \). This formula applies to our case where \( n = 150 \) and \( p = 0.4 \). Understanding how to apply this distribution effectively helps calculate exact probabilities in experiments and surveys.

The Role of Histogram Correction in Probability Approximation

Histogram, or continuity correction, refines the approximation of a discrete distribution by a continuous one, like binomial to normal. This correction is crucial when using the central limit theorem. The central limit theorem helps approximate the sum of a large number of independent and identically distributed random variables. However, it treats them as continuous variables.
A discrete random variable can assume specific values only (like whole numbers), but a continuous one fills the gaps in between. Therefore, when approximating, adjustments are needed. For a single value, such as \( S_n = 60 \), you correct it to a range \(59.5 < S_n < 60.5\) for the normal distribution.
After correcting the range, calculate the probability by finding the area under the normal curve between these bounds. This small adjustment to the interval improves accuracy, which in the original problem changed the approximation from essentially zero to a more realistic 0.076.

Exploring Normal Distribution

A normal distribution is a continuous probability distribution. It's often used because many psychological and natural phenomena tend to be normally distributed. It is famously bell-shaped and symmetric.
Characteristics include:

• Mean (bc): The peak and the center of the distribution.
• Standard deviation (c3): Measures the spread of the distribution.

In the problem, after transforming binomial data into normal form via the central limit theorem, it was assumed to approximate results due to the large \( n \) (150). Key to a normal distribution is its flexible application in situations with large sample sizes, ensuring figures become predictable. The central limit theorem states that, irrespective of a distribution's shape, sample means approach a normal distribution as \( n \) grows larger. This property allows transitions from discrete to continuous evaluations.

Importance of Standard Deviation

Standard deviation is a statistical measure that quantifies the dispersion of a dataset relative to its mean. In terms of probability, it shows how much variation there is from the average (mean) value.
Key aspects of the standard deviation include:

• Small standard deviation: Data points tend to be close to the mean.
• Large standard deviation: Data points are spread out over a larger range of values.

For the binomial distribution in our exercise, the standard deviation is calculated using:\[
3\sigma = \sqrt{n \times p \times (1-p)}\]This gives insight into how much the number of successes varies from one set of trials to another. In estimating probabilities using a normal distribution, it helps determine the z-scores (standardized scores) necessary for computations. In context, this measure helps determine more accurate probabilities by adjusting for variances within data.