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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

Short Answer

Expert verified
a. Approximately 0.0853 b. Approximately 0.7554 c. Approximately 0.0572

Step by step solution

01

Understanding the Problem

This problem involves a binomial random variable, with parameters \(n=200\) and \(p=0.4\). We will approximate the required probabilities using the normal distribution as a result of the Central Limit Theorem.
02

Finding Mean and Standard Deviation

For a binomial distribution, the mean \(\mu\) is given by \(np\) and the standard deviation \(\sigma\) by \(\sqrt{np(1-p)}\). Calculate these values: \(\mu = 200 \times 0.4 = 80\), and \(\sigma = \sqrt{200 \times 0.4 \times 0.6} \approx 6.928\).
03

Applying Continuity Correction

As we approximate using a continuous distribution, apply continuity correction to each probability: - For \(P(X \leq 70)\), use \(P(X \leq 70.5)\). - For \(P(70 < X < 90)\), use \(P(70.5 < X < 89.5)\). - For \(P(X = 80)\), use \(P(79.5 < X < 80.5)\).
04

Standardizing the Variable

Convert the corrected probabilities to standard normal variables (\(Z\)): - For \(P(X \leq 70.5)\), calculate \(Z_{70.5} = \frac{70.5 - 80}{6.928} \approx -1.37\). - For \(P(70.5 < X < 89.5)\), calculate \(Z_{70.5}\) and \(Z_{89.5} = \frac{89.5 - 80}{6.928} \approx 1.37\). - For \(P(79.5 < X < 80.5)\), calculate \(Z_{79.5} = \frac{79.5 - 80}{6.928} \approx -0.072\) and \(Z_{80.5} = \frac{80.5 - 80}{6.928} \approx 0.072\).
05

Using the Standard Normal Table

With the calculated \(Z\) values, use the standard normal (Z) table to find probabilities:- \(P(Z \leq -1.37)\) corresponds to \(P(X \leq 70)\).- \(P(-1.37 < Z < 1.37)\) by \(P(70.5 < X < 89.5)\).- \(P(-0.072 < Z < 0.072)\) by \(P(X = 80)\).
06

Conclusion of the Probabilities

Based on the Z values: - \(P(X \leq 70)\) is approximately 0.0853.- \(P(70 < X < 90)\) is approximately 0.7554.- \(P(X = 80)\) is approximately 0.0572.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is an important concept in statistics and is particularly useful in the field of probability. It tells us that the distribution of the sum (or average) of a large number of independent, identically distributed variables, regardless of the original distribution, approaches a normal distribution as the sample size becomes large enough.
This is a foundational concept because it allows us to approximate distributions that might not initially be normal.In the context of our binomial example with parameters \(n = 200\) and \(p = 0.4\), the CLT helps us to use a normal distribution for approximation. Given the large number of trials (\(n = 200\)), the binomial distribution can be effectively approximated by a normal distribution. This is because, as per the CLT, the sum of a large number of binomial random variables forms a bell-shaped curve, which aligns with a normal distribution.
Continuity Correction
In statistics, continuity correction is used when we approximate a discrete distribution, like the binomial distribution, with a continuous distribution, such as the normal distribution. Since a normal distribution is continuous, it doesn't have separate, individual values like discrete distributions do. Instead, it covers a range of values.For instance, to approximate a probability involving an exact number of successes in a binomial distribution, say \(P(X \leq 70)\), continuity correction helps adjust it to fit within the continuous framework. We convert this to \(P(X \leq 70.5)\) to account for the space between discrete numbers.
This small adjustment improves the accuracy of our approximation, bridging the gap between discrete step functions and continuous curves.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. This particular distribution is fundamental in statistics as it provides a reference point for standardizing other normal distributions. When we convert our values to the standard normal distribution, we can utilize the standard normal (or Z) table to find probabilities. This conversion is essential for comparing and calculating probabilities across different normal distributions, including ours. By transforming our binomial variable into this standard form, we can easily determine corresponding probabilities using the Z-table.
The concept of using a standard normal distribution simplifies many statistical tasks, as it provides a common scale for comparison.
Standardization of Variables
Standardization is the process of converting a variable into a common scale. This involves subtracting the mean from the variable and dividing by the standard deviation, represented in the formula:\[ Z = \frac{X - \mu}{\sigma} \]Where \(X\) is the value to be standardized, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. This process transforms our variable into a standard normal variable (Z), which has a mean of 0 and a standard deviation of 1.In this specific case, standardization simplifies the comparison of different data points and allows us to use the standard normal table to look up probabilities. For instance, converting a binomial outcome like \(X = 70.5\) to a Z-score allows us to assess its likelihood within the standard normal distribution framework, helping recognize how extreme or typical this value is.

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