/*! 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 26 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 to find probabilities and compare with exact binomial probabilities.

Step by step solution

01

Calculate Mean and Standard Deviation

For a binomial distribution, the mean \(\mu\) is given by \(np\) and the standard deviation \(\sigma\) is given by \(\sqrt{np(1-p)}\). Calculate these values for each \(p = 0.5, 0.6, 0.8\) for \(n = 25\).
02

Apply Normal Approximation

The normal approximation is used to approximate the binomial distribution. The continuity correction is applied by adjusting the boundaries of the interval by 0.5. Use the formula \[P(a \leq X \leq b) \approx P\left(\frac{a-0.5 - \mu}{\sigma} \leq Z \leq \frac{b+0.5 - \mu}{\sigma}\right)\]Determine \(Z\) for each probability interval with continuity correction for each value of \(p\).
03

Determine Z-scores and Find Probabilities

Calculate the Z-scores for each adjusted interval using the previously calculated \(\mu\) and \(\sigma\). Use the standard normal distribution table or a calculator to find the approximate probabilities.
04

Calculate Exact Binomial Probabilities

For the exact probabilities, use the binomial probability formula: \[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\]Calculate the sum of probabilities for each interval using this formula and compare it to 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
The binomial distribution is fundamental in probability and statistics. It models scenarios where there are exactly two possible outcomes for a fixed number of trials. Each trial is independent, and the probability of success remains constant across trials. For example, flipping a coin multiple times, where each flip results in either heads or tails, fits this distribution. The main parameters are:
  • \(n\): The total number of trials
  • \(p\): The probability of success on each trial
The mean \(\mu\) is calculated by \(np\), and variance is expressed as \(np(1-p)\). Understanding these parameters is crucial for calculations involving binomial probabilities, like mean and standard deviation.
Continuity Correction
When utilizing normal approximation for a binomial distribution, applying a continuity correction ensures a more accurate estimate. This adjustment accounts for the fact that the binomial distribution is discrete, while the normal distribution is continuous.To apply the continuity correction, adjust the endpoints of the interval of interest by 0.5. This small tweak bridges the gap between a discrete and a continuous model. For example, for a probability such as \(P(15 \leq X \leq 20)\), adjust the limits to \(14.5 \leq X \leq 20.5\). This step is pivotal because it refines the approximation by aligning more closely with the discrete nature of the data.
Z-scores
Z-scores transform a value from a normal distribution into a value on the standard normal distribution, which has a mean of 0 and a standard deviation of 1. By calculating \(Z\)-scores, different normal distributions become comparable, enabling probability lookups using a standard normal table.The formula for a Z-score is:\[Z = \frac{X - \mu}{\sigma}\]where \(X\) is the value from the original distribution, \(\mu\) is the mean, and \(\sigma\) is the standard deviation of the distribution. For intervals like \(P(a \leq X \leq b)\), compute the Z-scores using endpoints adjusted with continuity correction, hence, \(a-0.5\) and \(b+0.5\). This facilitates finding probabilities by searching standard normal distribution tables or using statistical software.
Probability Calculations
Probability calculations form the core of understanding distributions and making future predictions based on past data. When employing normal approximation, the key steps involve determining limits, calculating Z-scores post-continuity correction, and finally finding corresponding probabilities.Use the standard normal distribution table, where Z-scores correlate to cumulative probabilities. For each computed Z-score from limits like \(P(a \leq X \leq b)\), extract the probability. Deduct or add as necessary to solve for each interval.Comparatively, for exact binomial probabilities, apply the formula:\[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\]Calculate probabilities for the upper and lower bounds, then sum over these integer values for the exact result. Comparing exact and approximated probabilities show the approximation's efficiency, especially when \(n\) is large and \(p\) is not extreme (close to 0 or 1).

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Most popular questions from this chapter

When circuit boards used in the manufacture of compact dise players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly 10 defectives in the batch?

The following failure time observations (1000s of hours) resulted from accelerated life testing of 16 integrated circuit chips of a certain type: \(\begin{array}{rrrrrr}82.8 & 11.6 & 359.5 & 502.5 & 307.8 & 179.7 \\ 242.0 & 26.5 & 244.8 & 304.3 & 379.1 & 212.6 \\ 229.9 & 558.9 & 366.7 & 204.6 & & \end{array}\) Use the corresponding percentiles of the exponential distribution with \(\lambda=1\) to construct a probability plot. Then explain why the plot assesses the plausibility of the sample having been generated from any exponential distribution.

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985: 519-522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to \(\mathrm{AC}\) voltage. The values of the parameters depend on the voltage and temperature; suppose \(\alpha=2.5\) and \(\beta=200\) (values suggested by data in the article). a. What is the probability that a specimen's lifetime is at most 250 ? Less than 250 ? More than 300 ? b. What is the probability that a specimen's lifetime is between 100 and 250 ? c. What value is such that exactly \(50 \%\) of all specimens have lifetimes exceeding that value?

Mopeds (small motorcycles with an engine capacity below \(50 \mathrm{~cm}^{3}\) ) are very popular in Europe because of their mobility, ease of operation, and low cost. The article "Procedure to Verify the Maximum Speed of Automatic Transmission Mopeds in Periodic Motor Vehicle Inspections" ( \(J\). of Automobile Engr., 2008: 1615-1623) described a rolling bench test for determining maximum vehicle speed. A normal distribution with mean value \(46.8 \mathrm{~km} / \mathrm{h}\) and standard deviation \(1.75 \mathrm{~km} / \mathrm{h}\) is postulated. Consider randomly selecting a single such moped. a. What is the probability that maximum speed is at most \(50 \mathrm{~km} / \mathrm{h}\) ? b. What is the probability that maximum speed is at least \(48 \mathrm{~km} / \mathrm{h}\) ? c. What is the probability that maximum speed differs from the mean value by at most \(1.5\) standard deviations?

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