/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Let \(X\) have a binomial distri... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) have a binomial distribution with parameters \(n=25\) and \(p\). Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases \(p=.5, .6\), and \(.8\) and compare to the exact probabilities calculated from Appendix Table A.1. a. \(P(15 \leq X \leq 20)\) b. \(P(X \leq 15)\) c. \(P(20 \leq X)\)

Short Answer

Expert verified
Use normal approximation with continuity correction and compare results with exact values from Table A.1 for each probability case with \( p = 0.5, 0.6, \text{ and } 0.8 \).

Step by step solution

01

Calculate Mean and Standard Deviation

For a binomial distribution, the mean (\( \mu \)) and standard deviation (\( \sigma \)) are calculated as follows: \( \mu = n \times p \) and \( \sigma = \sqrt{n \times p \times (1-p)} \). We'll calculate these for each \( p \) value.
02

Normal Approximation with Continuity Correction for (a)

For event \( P(15 \leq X \leq 20) \), use continuity correction: \( P(14.5 < X < 20.5) \). Convert to the normal distribution using \( Z \) scores: \( Z = \frac{X - \mu}{\sigma} \). Calculate probabilities for each \( p \), using the cumulative standard normal distribution tables or a calculator.
03

Normal Approximation with Continuity Correction for (b)

For event \( P(X \leq 15) \), use continuity correction: \( P(X < 15.5) \). Convert \( X = 15.5 \) to \( Z \) score using \( Z = \frac{X - \mu}{\sigma} \) for each \( p \), and find the probability using standard normal tables.
04

Normal Approximation with Continuity Correction for (c)

For event \( P(20 \leq X) \), use continuity correction: \( P(X > 19.5) \). Calculate \( Z \) score for \( X = 19.5 \) using \( Z = \frac{X - \mu}{\sigma} \) for each \( p \), and find the probability using 1 minus the cumulative standard normal distribution value.
05

Compare with Exact Probabilities

Look up exact probabilities in Appendix Table A.1 for each case with \( p = 0.5, 0.6, \text{ and } 0.8 \) and compare with the approximated probabilities to see the accuracy of the normal approximation.

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Key Concepts

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

Binomial Distribution
A binomial distribution is a specific probability distribution that applies to discrete random events. It models situations where there are two possible outcomes, often referred to as "success" and "failure." The key parameters are:
  • The number of trials (\( n \)). This is how many times the experiment is conducted.
  • The probability of success in each trial (\( p \)). This value should be constant for each trial.
In the exercise case, we have 25 trials, meaning \( n = 25 \). You also have different probabilities of success for each scenario: \( p = 0.5 \), \( p = 0.6 \), and \( p = 0.8 \). The random variable \( X \) represents the number of successes in these 25 trials. Binomial distributions are often used to model real-world phenomena like the flipping of a coin or true/false assessments in statistics tests.
It's important to note that when working with binomial distributions, each trial is independent, and the probability for success remains unchanged throughout the process.
Continuity Correction
When using a continuous distribution like the normal distribution to approximate a discrete one like the binomial distribution, a continuity correction is applied. This adjustment accounts for the fact that discrete values are being approximated by a smooth, continuous curve.To carry out the continuity correction, you adjust your range by 0.5 units. For example:
  • For an inequality like \( P(15 \leq X \leq 20) \), you modify it to \( P(14.5 < X < 20.5) \).
  • For \( P(X \leq 15) \), it changes to \( P(X < 15.5) \).
  • And for \( P(20 \leq X) \), it becomes \( P(X > 19.5) \).
This small adjustment improves the accuracy of the normal approximation by better aligning the area under the curve to match the discrete probabilities of the binomial distribution. It plays a critical role when making the approximation more precise, especially when the number of trials, \( n \), isn't extremely large.
Standard Deviation
Standard deviation, a fundamental concept in statistics, measures the amount of variation or dispersion of a set of values. When dealing with a binomial distribution, standard deviation helps us understand the spread of possible outcomes based on given parameters.The formula for the standard deviation of a binomial distribution is:
  • \( \sigma = \sqrt{n \times p \times (1-p)} \)
Where:
  • \( n \) is the number of trials.
  • \( p \) is the probability of success on each trial.
  • \( 1-p \) represents the probability of failure on each trial.
The standard deviation tells us how much the results can deviate from the expected number of successes, \( \mu \). A larger standard deviation indicates more variability in the number of successes across multiple trials, while a smaller value points to more consistency.
Cumulative Standard Normal Distribution
The cumulative standard normal distribution is a way to calculate probabilities for the normal distribution, which is a continuous probability distribution. This cumulative form gives the probability that a standard normal random variable is less than or equal to a specified value.For a binomial distribution approximated as a normal distribution, you first convert the original values into Z-scores using the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here:
  • \( X \) is the value you're considering.
  • \( \mu \) is the mean of the distribution.
  • \( \sigma \) is the standard deviation.
Once you have the Z-score, you refer to Z-tables (or use technology like calculators) to find the probability that a normally-distributed variable is less than this Z-value. This helps us understand chances of results staying within a particular range in a standard, bell-shaped curve context. Using these cumulative tables is crucial for deriving probabilities when approximating binomial distributions with a normal distribution and applying continuity corrections.

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Most popular questions from this chapter

Stress is applied to a 20 -in. steel bar that is clamped in a fixed position at each end. Let \(Y=\) the distance from the left end at which the bar snaps. Suppose \(Y / 20\) has a standard beta distribution with \(E(Y)=10\) and \(V(Y)=\frac{100}{7}\). a. What are the parameters of the relevant standard beta distribution? b. Compute \(P(8 \leq Y \leq 12)\). c. Compute the probability that the bar snaps more than 2 in. from where you expect it to.

Suppose the force acting on a column that helps to support a building is a normally distributed random variable \(X\) with mean value \(15.0\) kips and standard deviation \(1.25\) kips. Compute the following probabilities by standardizing and then using Table A.3. a. \(P(X \leq 15)\) b. \(P(X \leq 17.5)\) c. \(P(X \geq 10)\) d. \(P(14 \leq X \leq 18)\) e. \(P(|X-15| \leq 3)\)

A system consists of five identical components connected in series as shown: As soon as one component fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with \(\lambda=.01\) and that components fail independently of one another. Define events \(A_{i}=\) \(\\{i\) th component lasts at least \(t\) hours \(\\}, i=1, \ldots, 5\), so that the \(A_{i} \mathrm{~s}\) are independent events. Let \(X=\) the time at which the system fails-that is, the shortest (minimum) lifetime among the five components. a. The event \(\\{X \geq t\\}\) is equivalent to what event involving \(A_{1}, \ldots, A_{5}\) ? b. Using the independence of the \(A_{i}{ }^{\prime}\) s, compute \(P(X \geq t)\). Then obtain \(F(t)=P(X \leq t)\) and the pdf of \(X\). What type of distribution does \(X\) have? c. Suppose there are \(n\) components, each having exponential lifetime with parameter \(\lambda\). What type of distribution does \(X\) have?

Vehicle speed on a particular bridge in China can be modeled as normally distributed ("Fatigue Reliability Assessment for Long-Span Bridges under Combined Dynamic Loads from Winds and Vehicles" \(J\). of Bridge Engr., 2013: 735-747). a. If \(5 \%\) of all vehicles travel less than \(39.12 \mathrm{~m} / \mathrm{h}\) and \(10 \%\) travel more than \(73.24 \mathrm{~m} / \mathrm{h}\), what are the mean and standard deviation of vehicle speed? [Note: The resulting values should agree with those given in the cited article.] b. What is the probability that a randomly selected vehicle's speed is between 50 and \(65 \mathrm{~m} / \mathrm{h}\) ? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of \(70 \mathrm{~m} / \mathrm{h}\) ?

The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 70 and standard deviation 3 . a. If a specimen is acceptable only if its hardness is between 67 and 75 , what is the probability that a randomly chosen specimen has an acceptable hardness? b. If the acceptable range of hardness is \((70-c, 70+c)\), for what value of \(c\) would \(95 \%\) of all specimens have acceptable hardness? c. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is indepen-dently determined, what is the expected number of acceptable specimens among the ten? d. What is the probability that at most eight of ten independently selected specimens have a hardness of less than 73.84?

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