/*! 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 27 Consider a binomial experiment w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a binomial experiment with \(n=20\) and \(p=.70\) a. \(\quad\) Compute \(f(12)\) b. Compute \(f(16)\) c. \(\quad\) Compute \(P(x \geq 16)\) d. Compute \(P(x \leq 15)\) e. Compute \(E(x)\) f. Compute \(\operatorname{Var}(x)\) and \(\sigma\).

Short Answer

Expert verified
\(f(12) \approx 0.114\), \(f(16) \approx 0.130\), \(P(x \geq 16) \approx 0.2211\), \(P(x \leq 15) \approx 0.7789\), \(E(x) = 14\), \(\operatorname{Var}(x) = 4.2\), \(\sigma \approx 2.05\).

Step by step solution

01

Understand the Problem

We are given a binomial experiment with the number of trials \(n=20\) and the probability of success \(p=0.70\). Binomial probabilities are calculated using the formula:\[ f(x) = \binom{n}{x} p^x (1-p)^{n-x} \]where \(\binom{n}{x}\) is the binomial coefficient.
02

Compute \(f(12)\)

We need to compute the probability of getting exactly 12 successes:\[ f(12) = \binom{20}{12} (0.70)^{12} (0.30)^{8} \]First, find \(\binom{20}{12} = \frac{20!}{12!(20-12)!}\). Calculating directly gives:\[ \binom{20}{12} = 125970 \]Therefore:\[ f(12) = 125970 \times (0.70)^{12} \times (0.30)^{8} \approx 0.114 \]
03

Compute \(f(16)\)

Compute the probability of getting exactly 16 successes:\[ f(16) = \binom{20}{16} (0.70)^{16} (0.30)^{4} \]Calculate \(\binom{20}{16} = \binom{20}{4} = 4845\). Thus:\[ f(16) = 4845 \times (0.70)^{16} \times (0.30)^{4} \approx 0.130 \]
04

Compute \(P(x \geq 16)\)

This is the sum of probabilities from 16 to 20, so calculate:\[ P(x \geq 16) = f(16) + f(17) + f(18) + f(19) + f(20) \]We use the binomial probability formula for each and sum them:- \(f(17) \approx 0.068\)- \(f(18) \approx 0.020\)- \(f(19) \approx 0.003\)- \(f(20) \approx 0.0001\)Thus:\[ P(x \geq 16) = 0.130 + 0.068 + 0.020 + 0.003 + 0.0001 = 0.2211 \]
05

Compute \(P(x \leq 15)\)

To find this probability, use the complement rule:\[ P(x \leq 15) = 1 - P(x \geq 16) \]Substituting the result from Step 4:\[ P(x \leq 15) = 1 - 0.2211 = 0.7789 \]
06

Compute \(E(x)\)

The expected value for a binomial distribution is given by:\[ E(x) = np \]So:\[ E(x) = 20 \times 0.70 = 14 \]
07

Compute \(\operatorname{Var}(x)\) and \(\sigma\)

The variance of a binomial distribution is:\[ \operatorname{Var}(x) = np(1-p) \]Thus:\[ \operatorname{Var}(x) = 20 \times 0.70 \times 0.30 = 4.2 \]The standard deviation \(\sigma\) is:\[ \sigma = \sqrt{\operatorname{Var}(x)} = \sqrt{4.2} \approx 2.05 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Probability in Binomial Distribution
Probability is a measure of how likely an event is to happen. In a binomial distribution, probabilities are calculated for a fixed number of trials, each with two possible outcomes: success or failure. For example, when flipping a coin, getting heads can be considered a success while tails would be a failure.
The formula to calculate the probability of exactly "x" successes in "n" trials is: \[ f(x) = \binom{n}{x} p^x (1-p)^{n-x} \] Here, \( p \) represents the probability of success, and the term \( \binom{n}{x} \) is the binomial coefficient, which tells us the number of ways we can pick "x" successes from "n" trials.
In our exercise, the probability of achieving exactly 12 successes out of 20 trials is determined by calculating \( f(12) \), which results in approximately 0.114. Similarly, the probability of having exactly 16 successes, \( f(16) \), is approximately 0.130. This illustrates how the binomial distribution provides a structured way to determine the likelihood of various outcomes.
Expected Value: The Mean of a Binomial Distribution
The expected value (or mean) of a binomial distribution provides insight into what the "average" outcome would be if the experiment was repeated many times. It is computed using the formula: \[ E(x) = np \] where "n" is the number of trials and "p" is the probability of success in each trial.
In our context, with \( n = 20 \) trials and a success probability of \( p = 0.70 \), the expected value is: \[ E(x) = 20 \times 0.70 = 14 \] This means, on average, you can expect about 14 successes in 20 trials. This average helps set expectations when observing outcomes in real-world scenarios where the binomial model is applicable.
The expected value is a central concept, as it serves as the balance point in a distribution, providing a simple understanding of the typical outcome.
Variance in Binomial Distribution: Measuring Dispersion
Variance measures the spread or dispersion of a set of values. In a binomial distribution, it tells us how much the number of successes deviates from the expected value. The formula used is: \[ \operatorname{Var}(x) = np(1-p) \] where "n" is the number of trials, and "p" is the probability of success.
For our experiment, the variance is calculated as: \[ \operatorname{Var}(x) = 20 \times 0.70 \times 0.30 = 4.2 \] This result indicates the variability around the expected value of 14 successes. The variance informs us how much actual results can differ from this expected average.
A higher variance indicates more spread-out results, whereas a lower variance implies results are more tightly clustered around the expected value.
Standard Deviation: Interpreting Variation
The standard deviation is the square root of the variance and it provides a way to understand the average distance values are from the mean. In a binomial distribution, it is computed as: \[ \sigma = \sqrt{\operatorname{Var}(x)} \] For our binomial experiment, the standard deviation is: \[ \sigma = \sqrt{4.2} \approx 2.05 \] This value indicates that the number of successes in the trials typically varies by around 2.05 from the mean of 14 successes.
It's an important concept in statistics because it not only considers variance but also translates it into understandable units. While variance provides the spread in squared units, standard deviation brings it back to the original units of the data, making it easier to interpret.
This measure helps in assessing consistency in data. For instance, a smaller standard deviation would mean the outcomes are generally closer to the mean.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Blackjack, or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards. Face cards (jacks, queens, and kings) and tens have a point value of \(10 .\) Aces have a point value of 1 or \(11 .\) A 52 -card deck contains 16 cards with a point value of 10 (jacks, queens, kings, and tens) and four aces. a. What is the probability that both cards dealt are aces or 10 -point cards? b. What is the probability that both of the cards are aces? c. What is the probability that both of the cards have a point value of \(10 ?\) d. \(A\) blackjack is a 10 -point card and an ace for a value of \(21 .\) Use your answers to parts (a), (b), and (c) to determine the probability that a player is dealt blackjack. (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts (a), (b), and (c) can be combined to answer this question.)

When a new machine is functioning properly, only \(3 \%\) of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. a. Describe the conditions under which this situation would be a binomial experiment. b. Draw a tree diagram similar to Figure 5.3 showing this problem as a two- trial experiment. c. How many experimental outcomes result in exactly one defect being found? d. Compute the probabilities associated with finding no defects, exactly one defect, and two defects.

A new automated production process averages 1.5 breakdowns per day. Because of the cost associated with a breakdown, management is concerned about the possibility of having three or more breakdowns during a day. Assume that breakdowns occur randomly, that the probability of a breakdown is the same for any two time intervals of equal length, and that breakdowns in one period are independent of breakdowns in other periods. What is the probability of having three or more breakdowns during a day?

The Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United States. For 15 restaurants located in Boston, the average price of a dinner, including one drink and tip, was \(\$ 48.60 .\) You are leaving for a business trip to Boston and will eat dinner at three of these restaurants. Your company will reimburse you for a maximum of \(\$ 50\) per dinner. Business associates familiar with these restaurants have told you that the meal cost at one-third of these restaurants will exceed \(\$ 50 .\) Suppose that you randomly select three of these restaurants for dinner. a. What is the probability that none of the meals will exceed the cost covered by your company? b. What is the probability that one of the meals will exceed the cost covered by your company? c. What is the probability that two of the meals will exceed the cost covered by your company? d. What is the probability that all three of the meals will exceed the cost covered by your company?

To perform a certain type of blood analysis, lab technicians must perform two procedures. The first procedure requires either one or two separate steps, and the second procedure requires either one, two, or three steps. a. List the experimental outcomes associated with performing the blood analysis. b. If the random variable of interest is the total number of steps required to do the complete analysis (both procedures), show what value the random variable will assume for each of the experimental outcomes.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.