Problem 19 - Calculus for Biology and Medicine

Question

Suppose \(S_{n}\) is binomially distributed with parameters \(n=\) 200 and \(p=0.3 .\) Use the central limit theorem to find an approximation for \(P\left(99 \leq S_{n} \leq 101\right)\) (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

Both approximations, with and without continuity correction, suggest a probability very close to 0, similar to the exact binomial calculation using technology.

Step 1: Understand the Problem

We are given a binomial distribution with parameters \( n = 200 \) and \( p = 0.3 \). We need to approximate the probability \( P(99 \leq S_n \leq 101) \) using the central limit theorem (CLT) both without and with histogram correction.

Step 2: Define the Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) is given by \( \mu = np \). Here, \( \mu = 200 \times 0.3 = 60 \). The standard deviation \( \sigma \) is given by \( \sigma = \sqrt{np(1-p)} \). So, \( \sigma = \sqrt{200 \times 0.3 \times 0.7} \approx 6.48 \).

Step 3: Use Central Limit Theorem without Histogram Correction

According to CLT, we can approximate the binomial distribution with a normal distribution. Convert \( S_n \) to a standard normal variable \( Z \) as \( Z = \frac{X - \mu}{\sigma} \). Calculate \( P(99 \leq S_n \leq 101) \) as follows:\[P(99 \leq S_n \leq 101) \approx P\left( \frac{99-60}{6.48} \leq Z \leq \frac{101-60}{6.48} \right)\]This becomes:\[P(6.02 \leq Z \leq 6.33)\]

Step 4: Calculate Probability without Histogram Correction

Using standard normal distribution tables or technology:- \( Z=6.02 \) corresponds to a very high percentile(probability close to 1)- \( Z=6.33 \) is even higher (closer to 1)Thus, \( P(6.02 \leq Z \leq 6.33) \) is nearly 0, meaning the probability without correction is practically 0.

Step 5: Use Central Limit Theorem with Histogram Correction

For histogram correction (continuity correction), adjust the values by \( \pm 0.5 \):\[P(98.5 \leq S_n \leq 101.5) \approx P\left( \frac{98.5-60}{6.48} \leq Z \leq \frac{101.5-60}{6.48} \right)\]This becomes:\[P(5.93 \leq Z \leq 6.41)\]

Step 6: Calculate Probability with Histogram Correction

Using standard normal distribution tables or technology:- \( Z=5.93 \) is close to 1- \( Z=6.41 \) is even closer to 1Thus, \( P(5.93 \leq Z \leq 6.41) \) is also nearly 0, making the probability with correction practically 0.

Step 7: Exact Probability Calculation

Use a graphing calculator or statistical software to compute the exact binomial probabilities:- \( P(S_n = 99) \), \( P(S_n = 100) \), and \( P(S_n = 101) \) add up to the probability of \( P(99 \leq S_n \leq 101) \).The exact probability is the summation of these calculated values. Compare these with the nearly zero probabilities obtained from normal approximation methods.

Key Concepts

Binomial Distribution

The binomial distribution is a discrete probability distribution. It describes the number of successes in a fixed number of independent trials of a binary experiment. A binary experiment is one where there are only two possible outcomes, like flipping a coin where you can only get heads or tails.

• "Success" is defined as the outcome of interest. The probability of success on an individual trial is denoted by \( p \).
• The number of trials, denoted by \( n \), is fixed, and each trial is independent from the others.

In the given exercise, we have a binomial distribution with parameters \( n = 200 \) and \( p = 0.3 \) which means there are 200 trials in which the probability of success is 30%.
This type of distribution becomes important when calculating probabilities of a number of successes within the trials.

Normal Distribution Approximation

When dealing with a large number of trials, as with \( n = 200 \) in the exercise, the binomial distribution can be approximated by a normal distribution due to the Central Limit Theorem (CLT).
The CLT states that as the sample size becomes larger, the distribution of the sample mean will tend to be normally distributed. Here are a few key points:

• The mean \( \mu \) of the normal distribution is \( np \).
• The standard deviation \( \sigma \) is \( \sqrt{np(1-p)} \).

This allows us to use the normal distribution to approximate probabilities when \( n \) is large, simplifying the calculation process.However, since we are approximating a discrete distribution (binomial) using a continuous one (normal), we need to consider adjustments for better accuracy.

Continuity Correction

Continuity correction comes into play when a discrete distribution is approximated by a continuous distribution. This correction accounts for the differences between data types.
In the exercise, the continuity correction is applied by adjusting the values sought by \( \pm 0.5 \).
For example, instead of calculating \( P(99 \leq S_n \leq 101) \), we compute \( P(98.5 \leq S_n \leq 101.5) \).

• This adjustment helps to align the discrete binomial bar with the continuous normal curve.
• It improves the approximation by compensating for the discrete nature of the binomial distribution.

Using continuity correction often yields probabilities that are much closer to the exact probability of the discrete binomial distribution. This correction is particularly useful when \( n \) is not extremely large, as it enhances accuracy in those cases.