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Chapter 10: Problem 35

Suppose \(X\) has the Binomial \((n, p)\) distribution. Use the normal approximation to estimate the given probability. \(P(X<30)\) if \(n=64, p=1 / 2\)

Short Answer

Expert verified
The estimated probability \(P(X < 30)\) is approximately 0.3085.

Step by step solution

01

- Identify the distribution parameters

Given that the random variable \(X\) follows a Binomial distribution with parameters \(n = 64\) and \(p = \frac{1}{2}\).
02

- Calculate the mean and variance

The mean \(\mu\) of a Binomial \((n, p)\) distribution is given by \(\mu = np\). The variance \(\sigma^2\) is given by \(\sigma^2 = np(1 - p)\). Here, \(\mu = 64 \times \frac{1}{2} = 32\) and \(\sigma^2 = 64 \times \frac{1}{2} \times \frac{1}{2} = 16\).
03

- Convert to standard normal distribution

To use the normal approximation, convert the Binomial variable \(X\) to a standard normal variable \(Z\) using the transformation \(Z = \frac{X - \mu}{\sigma}\). Here, \(\sigma = \sqrt{16} = 4\). We need to find the probability \(P(X < 30)\). First, find the corresponding Z-value: \(Z = \frac{30 - 32}{4} = -\frac{2}{4} = -0.5\).
04

- Use standard normal distribution table

Use the standard normal distribution table to find \(P(Z < -0.5)\). The table shows that \(P(Z < -0.5) \approx 0.3085\).

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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 discrete probability distribution. It describes the number of successes in a fixed number of trials, each with the same probability of success. In our exercise, we have a Binomial distribution characterized by the parameters \( n \) and \( p \):
  • \( n \) is the total number of trials, which in this case is 64.
  • \( p \) is the probability of success on each trial, which is given as \( \frac{1}{2} \).
When using the Binomial distribution, it allows us to find the probability of observing a specific number of successes. However, when \( n \) is large, calculating these probabilities directly can become complex.
This is where the normal approximation comes in handy.
Mean and Variance Calculation
Calculating the mean and variance is a crucial step in using the normal approximation to the binomial distribution. For a binomial distribution with parameters \( n \) and \( p \):
  • The mean \( \mu \) is calculated using \( \mu = np \). In our example, \( \mu = 64 \times \frac{1}{2} = 32 \).
  • The variance \( \sigma^2 \) is calculated using \( \sigma^2 = np(1 - p) \). Here, \( \sigma^2 = 64 \times \frac{1}{2} \times \frac{1}{2} = 16 \).
  • The standard deviation \( \sigma \) is simply the square root of the variance, which is \( \sqrt{16} = 4 \).
These calculations are essential because they allow us to convert our binomially distributed variable into a standard normal variable.
Standard Normal Distribution
Using the standard normal distribution simplifies the process of finding probabilities for binomial distributions, especially with large \( n \). The transformation to the standard normal variable \( Z \) is given by:
  • \( Z = \frac{X - \mu}{\sigma} \)
    • where \( X \) is our binomial variable, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In this exercise, to find:
  • \( P(X < 30) \)
    • we first convert 30 to the corresponding \( Z \)-value:
  • \( Z = \frac{30 - 32}{4} = -0.5 \)
    • We then use the standard normal distribution table to find \( P(Z < -0.5) \).From the table:
  • \( P(Z < -0.5) \approx 0.3085\)
    • Thus, the normal approximation gives us \( P(X < 30) \approx 0.3085 \).
    This method saves a lot of computation time and is highly useful for large sample sizes.

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Most popular questions from this chapter

Assume that the random variable \(X\) is normally distributed. Use the given information to find the unknown parameter or parameters of the distribution. If \(\mathrm{SD}(X)=2\) and \(P(X \geq-1)=0.409\), find \(E(X)\)

A significance level \(\alpha\) and a tail of the standard normal distribution are given. Use the normal table to approximately determine the critical value. \(\alpha=0.01\), left tail

The number of people in the United States of any age \(x\) between \(l\) and 85 who died in 2000 can be approximated by the function \(^{22}\) $$ f(x)=1152.9 e^{0.051476 x} $$ a) Find a value \(k\) to make the function \(k f(x)\) a probability density function. b) Use your answer in part (a) to approximate the mean age among all who died in 2000 with age between 1 and 85 . c) Find the standard deviation of the ages of people who died in 2000 whose age was between 1 and \(85 .\)

Verify the two properties of a probability density function over the given interval. $$ f(x)=\frac{1}{3},[4,7] $$

Let \(X\) be a continuous random variable. The \(p\) th percentile of its distribution is defined to be the number \(t\) so that the probability that \(x\) is less than or equal to \(\ell\) is \(p \%\) $$ P(X \leq t)=\frac{p}{100} $$ For example, if \(X\) has the standard normal density, then $$ P(X \leq 1.28) \approx 90 \% $$ using Table 2 . Therefore, we say that \(1.28\) is the 90 th percentile for a standard normal distribution. Find the following percentiles for a standard normal distribution. a) 30 th percentile b) 50 th percentile c) \(95 \mathrm{th}\) percentile
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