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Suppose that \(Y\) has a binomial distribution with \(n=5\) and \(p=.10\) a. Use the Normal Approximation to Binomial Distribution applet to find exact and approximate values for \(P(Y \leq 1)\) b. The normal approximation is not particularly good. Why?

Short Answer

Expert verified
The exact \(P(Y \leq 1) = 0.91854\), with normal approximation \(\approx 0.9319\); not accurate due to small \(n\) and \(p\).

Step by step solution

01

Identify Binomial Distribution Parameters

The binomial distribution is characterized by parameters: the number of trials \(n = 5\) and the probability of success on each trial \(p = 0.10\). We aim to find \(P(Y \leq 1)\), where \(Y\) follows a binomial distribution with these parameters.
02

Calculate Exact Binomial Probability

The exact probability can be calculated using the binomial probability formula: \[P(Y \leq 1) = P(Y = 0) + P(Y = 1).\]The formula for a binomial probability is \[P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Calculate \(P(Y = 0)\) and \(P(Y = 1)\):\[P(Y = 0) = \binom{5}{0} (0.10)^0 (0.90)^5 = 0.59049,\]\[P(Y = 1) = \binom{5}{1} (0.10)^1 (0.90)^4 = 0.32805.\]Thus,\[P(Y \leq 1) = 0.59049 + 0.32805 = 0.91854.\]
03

Determine Normal Approximation Parameters

Using the normal approximation to the binomial distribution, the mean (\(\mu\)) and variance (\(\sigma^2\)) of \(Y\) are given by:\[\mu = np = 5 \times 0.10 = 0.5,\]\[\sigma^2 = np(1-p) = 5 \times 0.10 \times 0.90 = 0.45.\]The standard deviation is \(\sigma = \sqrt{0.45} \approx 0.671\).
04

Use Continuity Correction for Normal Approximation

To better approximate the binomial distribution using the normal distribution, apply the continuity correction. We calculate:\[P(Y \leq 1) \approx P(Y < 1.5).\]
05

Calculate Approximate Normal Probability

Convert \(Y < 1.5\) to a standard normal variable \(Z\):\[Z = \frac{1.5 - \mu}{\sigma} = \frac{1.5 - 0.5}{0.671} \approx 1.49.\]Using the standard normal distribution table, find \(P(Z < 1.49) \approx 0.9319\).
06

Analyze the Accuracy of the Normal Approximation

The approximate probability of \(P(Y \leq 1)\) using normal distribution is \(0.9319\), whereas the exact probability is \(0.91854\). The difference is due to the small \(n\), and the success probability \(p\) being far from 0.5, which makes the approximation less accurate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
The binomial distribution is a widely-used probability distribution in statistics, especially useful for modeling events with two possible outcomes, often termed as "success" and "failure." In any binomial experiment, the number of trials is fixed, denoted by \(n\), and each trial has the same probability of success, represented by \(p\). For our exercise, \(n = 5\) and \(p = 0.10\).

The goal is to calculate the probability of a certain number of successes within these trials. This can involve calculating the probability for one outcome (like \(P(Y = 1)\)) or a range of outcomes (like \(P(Y \leq 1)\)), as shown in the provided steps. The probabilities can directly be found using the binomial probability formula:
  • \(P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
Continuity Correction
In probability, when using the normal approximation to a discrete distribution like the binomial distribution, a continuity correction is needed. This is because normal distributions are continuous, while binomial distributions are discrete. The correction helps align the areas under the curve, leading to a more accurate approximation.

In our example, we are interested in finding \(P(Y \leq 1)\). The continuity correction suggests that instead of finding \(P(Y \leq 1)\) directly, we should approximate this as \(P(Y < 1.5)\). This involves "smoothing" the jump from one integer to the next by adding or subtracting 0.5. Continuity correction is especially critical when dealing with small \(n\) or extreme \(p\) values, improving approximation accuracy.
Standard Normal Distribution
The standard normal distribution is a special form of the normal distribution. It has a mean of 0 and a standard deviation of 1. It's useful because any normal distribution can be converted to a standard normal distribution by transforming the variables.

For transformation, you use the formula:
  • \(Z = \frac{X - \mu}{\sigma}\)
where \(\mu\) is the mean and \(\sigma\) is the standard deviation of the original distribution.

In our exercises, after using continuity correction, we transform \(Y < 1.5\) into a standard normal variable \(Z\). Calculating \(Z\) helps us use standard normal distribution tables (or software) to find probabilities associated with these transformed values, greatly simplifying the process.
Probability Approximation
Probability approximation is a technique used in statistics to estimate the probability of a random variable falling within a certain range. When dealing with the binomial distribution, direct probability calculation becomes cumbersome, especially for large sample sizes. Hence, we rely on approximations like the normal approximation.

In this task, the exact probability for a binomial event is compared with its approximate probability using the normal distribution. Approximations are generally quite reliable, especially when \(n\) is large and \(p\) is close to 0.5. However, when \(n\) is small or \(p\) is far from 0.5, as in \(n = 5\) and \(p = 0.10\), the accuracy diminishes.

Understanding which methods to use in different scenarios and why approximations may vary is crucial for statistical analysis and problem-solving.

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