Question

Let \(x\) be a binomial random variable with \(n=25\) and \(p=.3\) a. Is the normal approximation appropriate for this binomial random variable? b. Find the mean and standard deviation for \(x\). c. Use the normal approximation to find \(P(6 \leq x \leq 9)\). d. Use Table 1 in Appendix I to find the exact probability \(P(6 \leq x \leq 9)\). Compare the results of parts \(c\) and d. How close was your approximation?

Short answer

#Answer# a. The normal approximation might not be considered appropriate since np < 10 and nq > 10. b. Mean (μ) = 7.5 and Standard Deviation (σ) ≈ 2.29. c. Normal approximation probability P(6 ≤ x ≤ 9) ≈ 0.3869. d. Exact probability P(6 ≤ x ≤ 9) ≈ 0.3832. The normal approximation was reasonably close to the exact probability, with a difference of about 0.0037 or 0.37%.

Step-by-step solution

a. Checking the appropriateness of the normal approximation

To check the appropriateness of the normal approximation, we can use the rule of thumb: the normal approximation is considered appropriate if \(np \geq 10\) and \(nq \geq 10\). In this case, \(n=25\) and \(p=0.3\), so \(q=1-p=0.7\). Then, \(np = 25 \times 0.3 = 7.5\) and \(nq = 25 \times 0.7 = 17.5\). Since \(np < 10\) and \(nq > 10\), the normal approximation might not be considered appropriate. However, we can still proceed if it is reasonable to assume that the normal approximation is close enough to the binomial distribution.

b. Finding the mean and standard deviation

For a binomial random variable with parameters \(n\) and \(p\), the mean (\(\mu\)) and standard deviation (\(\sigma\)) can be calculated as follows: \(\mu = np\) and \(\sigma = \sqrt{npq}.\) Substituting the values, we get the mean \(\mu = 25 \times 0.3 = 7.5\) and the standard deviation \(\sigma = \sqrt{25 \times 0.3 \times 0.7} \approx 2.29.\)

c. Normal approximation to find \(P(6 \leq x \leq 9)\)

First, we need to convert the binomial random variable \(x\) to a standard normal random variable \(z\). We can use the formula \(z = \frac{x-\mu}{\sigma}\). Using the continuity correction, we will replace \(6\) with \(5.5\) and \(9\) with \(9.5\). So the probability \(P(6 \leq x \leq 9)\) becomes \(P(5.5 \leq x \leq 9.5)\). Now we can find the corresponding \(z\) scores for each \(x\) value: \(z_1 = \frac{5.5 - 7.5}{2.29} \approx -0.87\) \(z_2 = \frac{9.5 - 7.5}{2.29} \approx 0.87\) Now, we can find the approximate probability using the normal distribution table or a computer software to find \(P(-0.87 \leq z \leq 0.87) \approx 0.3869.\)

d. Finding exact probability and comparing the results

We will now find the exact binomial probability \(P(6 \leq x \leq 9)\) using the binomial formula, this can be done with the Table 1 in Appendix I or computer software. \(P(6 \leq x \leq 9) = P(x=6)+P(x=7)+P(x=8)+P(x=9) = \sum_{k=6}^{9} {25\choose k} (0.3)^k (0.7)^{25-k} \approx 0.3832.\) Now, we can compare the approximate probability from part (c) and the exact probability from part (d). The approximate probability is \(0.3869\) and the exact probability is \(0.3832\). Thus, the normal approximation was reasonably close to the exact probability, with a difference of about \(0.0037\) or 0.37%.

Key concepts

Binomial Random Variable

A binomial random variable represents the number of successes in a fixed number of independent trials, each with the same probability of success. It's one of the building blocks of probability theory, and understanding it opens the door to a wide range of real-world applications, from genetics to quality control.

Consider a scenario where you flip a fair coin 25 times and count the number of heads. Each flip is an independent event with two possible outcomes – head or tail. The number of heads that turn up in those 25 flips is a perfect example of a binomial random variable. Here are some key points you must remember about it:

• The trials are independent; the outcome of one trial doesn't affect the others.
• Each trial has only two possible outcomes – a 'success' or a 'failure'.
• The probability of success, denoted as p, remains constant for each trial.
• The binomial random variable represents the count of successes.

For a binomial process with n trials and a probability p of success in each trial, the variable is expressed as X ~ B(n, p), where X denotes the binomial random variable.

Binomial Distribution

The binomial distribution is a discrete probability distribution that describes the number of successes in a sequence of independent experiments. The classic example is a series of coin tosses, but it could be applied to any process with a fixed number of trials and a consistent probability of success.

When working with binomial distributions, two important parameters are the mean (μ) and the standard deviation (σ), which gives us an idea about the ‘center’ of the distribution and how spread out it is, respectively. The mean for a binomial distribution is calculated as μ = np, and the standard deviation as σ = √(npq),where q = 1 - p.

The calculation of exact probabilities can be cumbersome, especially for large n. This is where the normal approximation comes into play. Under certain conditions (when np and nq are sufficiently large), the binomial distribution can be approximated by the standard normal distribution, which simplifies calculations and allows for the use of standard normal tables or software.

Standard Normal Distribution

The standard normal distribution, also known as the z-distribution, is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. It's the go-to tool for statisticians and researchers when dealing with problems involving normal distributions.

In essence, the standard normal distribution serves as a reference table for comparing different normal distributions. Any normal distribution can be transformed into a standard normal distribution using the formula z = (x - μ) / σ, where x is a value from the original distribution, μ is the mean, and σ is the standard deviation. This process is known as 'standardizing'.

A significant advantage of the standard normal distribution is the availability of the cumulative distribution function, which provides probabilities for values less than or equal to a given z-score. This is particularly helpful when using the normal approximation to the binomial distribution, as it allows you to find probabilities for a range of values conveniently. A z-score essentially tells you how many standard deviations away from the mean a value is, which makes comparing different data sets much easier.