/*! 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 69 For a binomial probability distr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 69

For a binomial probability distribution, \(n=120\) and \(p=.60\). Let \(x\) be the number of successes in 120 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \leq 69)\) using the normal approximation. c. Find \(P(67 \leq x \leq 73)\) using the normal approximation.

Short Answer

Expert verified
a. Mean = 72, Standard Deviation 鈮� 6.93, b. P(x 鈮� 69) 鈮� 0.3336 using the normal approximation, c. P(67 鈮� x 鈮� 73) 鈮� 0.3199 using the normal approximation.

Step by step solution

01

Calculate Mean and Standard Deviation

The mean and standard deviation of a binomial distribution can be calculated using the formula:Mean (\( \mu \)) = np, Standard Deviation (\( \sigma \)) = \sqrt{np(1-p)}Here, n = 120 and p = 0.60. Therefore, the mean (\( \mu \)) = np =120*0.60 = 72 and the standard deviation (\( \sigma \)) = \sqrt{np(1-p)}= \sqrt{120 * 0.60 * 0.40} 鈮� 6.93.
02

Calculate P(x 鈮� 69) using Normal Approximation

For P(x 鈮� 69) using the Normal Approximation, first convert x to a z-score using the formula: z = (x - 渭) / 蟽. Here, x = 69, so z = (69 - 72) / 6.93 鈮� -0.43. Using a standard z-table, look up the area to the left of z = -0.43 to find the probability. Assume a z-table gives a probability of 0.3336 which corresponds to P(x 鈮� 69).
03

Calculate P(67 鈮� x 鈮� 73) using Normal Approximation

For P(67 鈮� x 鈮� 73), calculate two z-scores: z1 for x = 67 and z2 for x = 73. z1 = (67 - 72) / 6.93= -0.72 and z2 = (73 - 72) / 6.93 = 0.14. Using a z-table, find the probability for z1 and z2. Let's say the probabilities from the z-Table are 0.2358 and 0.5557 respectively. The probability P(67 鈮� x 鈮� 73) = probability of z2 - probability of z1 = 0.5557 - 0.2358 = 0.3199.

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 91影视!

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

According to a Gallup poll, \(92 \%\) of Americans believe in God (Time, June 20,2011 ). Suppose that this result is true for the current population of adult Americans. What is the probability that the number of adult Americans in a sample of 500 who believe in God is a. exactly 445 b. at least 450 c. 440 to 470

At Jen and Perry Ice Cream Company, a machine fills 1-pound cartons of Top Flavor ice cream. The machine can be set to dispense, on average, any amount of ice cream into these cartons. However, the machine does not put exactly the same amount of ice cream into each carton; it varies from carton to carton. It is known that the amount of ice cream put into each such carton has a normal distribution with a standard deviation of 18 ounce. The quality control inspector wants to set the machine such that at least \(90 \%\) of the cartons have more than 16 ounces of ice cream. What should be the mean amount of ice cream put into these cartons by this machine?

The lengths of 3 -inch nails manufactured on a machine are normally distributed with a mean of \(3.0\) inches and a standard deviation of \(.009\) inch. The nails that are either shorter than \(2.98\) inches or longer than \(3.02\) inches are unusable. What percentage of all the nails produced by this machine are unusable?

Lori just bought a new set of four tires for her car. The life of each tire is normally distributed with a mean of 45,000 miles and a standard deviation of 2000 miles. Find the probability that all four tires will last for at least 46,000 miles. Assume that the life of each of these tires is independent of the lives of other tires.

Determine the following probabilities for the standard normal distribution. a. \(P(-2.46 \leq z \leq 1.88)\) b. \(P(0 \leq z \leq 1.96)\) c. \(P(-2.58 \leq z \leq 0)\) d. \(P(z \geq .73)\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation
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.