Problem 95 - Applied Statistics and Probability for Engineers

Question

Suppose that \(X\) is a binomial random variable with \(n=200\) and \(p=0.4\). Approximate the following probabilities: (a) \(P(X \leq 70)\) (b) \(P(70

Step-by-Step Solution

Answer

(a) \( \approx 0.0853 \); (b) \( \approx 0.8294 \); (c) \( \approx 0.056 \).

Step 1: Understand the Binomial Random Variable

The exercise considers a binomial random variable, where we have a fixed number of trials, \( n = 200 \), and the probability of success in each trial is \( p = 0.4 \).

Step 2: Use Normal Approximation

Since \( n \) is large, we can use the normal approximation to approximate these binomial probabilities. We calculate the mean \( \mu = np \) and the standard deviation \( \sigma = \sqrt{np(1-p)} \). \( \mu = 200 \times 0.4 = 80 \), \( \sigma = \sqrt{200 \times 0.4 \times 0.6} = \sqrt{48} \approx 6.93 \).

Step 3: Apply Continuity Correction

For parts (a) and (b), we need to apply the continuity correction when using the normal distribution. For \( P(X \leq 70) \), use \( P(X \leq 70.5) \). For \( P(70 < X < 90) \), use \( P(70.5 < X < 89.5) \).

Step 4: Convert to Standard Normal Distribution

Convert the binomial values into standard normal (Z) values using the formula \( Z = \frac{X - \mu}{\sigma} \). For example, to find \( P(X \leq 70) \), calculate \( Z = \frac{70.5 - 80}{6.93} \approx -1.37 \).

Step 5: Find Probabilities Using Z-Table

Use the standard normal distribution table (Z-table) to find probabilities. (a) For \( P(X \leq 70) \), find \( P(Z \leq -1.37) \approx 0.0853 \). (b) For \( P(70 < X < 90) \), find \( P(Z < \frac{89.5 - 80}{6.93}) - P(Z < \frac{70.5 - 80}{6.93}) \approx P(Z < 1.37) - P(Z < -1.37) \approx 0.9147 - 0.0853 = 0.8294 \).

Step 6: Find Specific Probability for Part (c)

Using the normal approximation, calculate \( P(79.5 < X < 80.5) \). Convert to Z values: \( P\left(Z > \frac{79.5 - 80}{6.93}\right) - P\left(Z < \frac{80.5 - 80}{6.93}\right) \approx P(Z < 0.072) - P(Z < -0.072) \approx 0.528 - 0.472 = 0.056 \).

Key Concepts

Normal Approximation

When you're dealing with a binomial distribution with a large number of trials, like in our case with 200 trials, calculations can become cumbersome. That's where normal approximation steps in to make life easier. The essence of this concept is to use a normal distribution to estimate the binomial probabilities. This works particularly well when the sample size, denoted as \( n \), is large. In our problem, we calculated the mean \( \mu = np = 80 \) and the standard deviation \( \sigma = \sqrt{np(1-p)} \approx 6.93 \). A normal distribution with this mean and standard deviation can approximate our binomial distribution for probabilities like \( P(X \leq 70) \). This approach simplifies many computations where direct application of the binomial formula would be complex.

Continuity Correction

Continuity correction is a small adjustment applied when using a continuous distribution, like the normal distribution, to estimate probabilities of a discrete distribution, such as the binomial distribution.Since the normal distribution is continuous and the binomial is discrete, the continuity correction helps bridge this gap. It involves adjusting the discrete values by ±0.5 to better fit into the continuous curve.For instance, to find \( P(X \leq 70) \), we actually calculate \( P(X \leq 70.5) \). Similarly, for \( P(70 < X < 90) \), the calculation becomes \( P(70.5 < X < 89.5) \). Applying continuity correction ensures that our approximation using the normal curve is more accurate.

Standard Normal Distribution

Once you have your normal distribution approximation, it helps to convert it into the standard normal distribution, which has a mean of 0 and a standard deviation of 1. This "standardization" simplifies the process of finding probabilities.The conversion formula is \( Z = \frac{X - \mu}{\sigma} \). This formula allows you to translate any point on a normal distribution to its equivalent on the standard normal curve. For example, to compute the probability for \( P(X \leq 70) \), translate 70.5 into \( Z \)-value: \( Z = \frac{70.5 - 80}{6.93} \approx -1.37 \).This conversion is crucial as it enables the use of standard tools like Z-tables for finding probabilities, making the process efficient and unified.

Z-Table

The Z-table is an essential tool in statistics that shows the probabilities of a standard normal distribution up to a given \( Z \)-value. Once you have a \( Z \)-value, as we calculated for our problem, you can use the Z-table to find the corresponding probability.For instance, the probability for \( Z \leq -1.37 \) provides \( P(X \leq 70) \approx 0.0853 \). Similarly, when solving \( P(70 < X < 90) \), by using the Z-table, the difference in probabilities can be calculated: \( P(Z < 1.37) - P(Z < -1.37) \approx 0.9147 - 0.0853 = 0.8294 \).The Z-table effectively consolidates complex calculations into simple lookup operations, streamlining the process of finding probabilities in statistics.