Certainly! Here’s a transcription suitable for an educational website:---**Problem Statement:**62% of all Americans live in cities with a population greater than 100,000 people. If 41 Americans are randomly selected, find the probability that:a. Exactly 24 of them live in cities with a population greater than 100,000 people.b. At most 27 of them live in cities with a population greater than 100,000 people.c. At least 23 of them live in cities with a population greater than 100,000 people.d. Between 23 and 30 (including 23 and 30) of them live in cities with a population greater than 100,000 people.**Hint:**- [Hint]- [Video on Finding Binomial Probabilities](#)[Submit Question]---**Additional Information:**- The text uses binomial probability to determine the likelihood of a certain number of Americans living in large cities based on a random sample.- To solve these problems, use binomial probability formulas and consider the given percentage of 62%.**Helpful Resources:**A link to a video resource on finding binomial probabilities is provided for further explanation and detailed understanding.--- (Note: No graphs or diagrams are present in the image.)

Given : 62% of all Americans live in cities with population greater than 100,000 people . 41 Americans are randomly selected . Let, X be the number of Americans live in cities with population greater than 100,000 people . Here, X ~ Binomial ( n = 41 , p = 0.62) When the sample size nn is large enough, and/or when p is close to 1/2, then X is approximately normally distributed. Since , n > 30 . It approximate to normal distribution . It is determined by two parameters: the population proportion of success, the sample size (or number of trials) The mean of the binomial distribution is n⋅p and its standard deviation is np(1-p) Therefore , The population mean is computed as: and the population standard deviation is computed as: (a) Exact 24 of them live in all cities . P (X = 24) = ? Normal approximation : P(x -0 .5 < X < x+0.5) = P (23.5 < X < 24.5) = =P23.5-25.423.108≤Z≤24.5-25.423.108=P(-0.6177 ≤Z≤-0.29601)=Pr(Z ≤-0.2960) - Pr(Z ≤ -0.6177)=0.3821 - 0.2676=0.1145 (b) At most 27 : (c) At least 23 : (d) Between 23 to 30 :

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62% of all Americans live in cities with population greater than 100,000 people. If 41 Americans are randomly selected, find the probability that a. Exactly 24 of them live in cities with population greater than 100,000 people. b. At most 27 of them live in cities with population greater than 100,000 people. c. At least 23 of them live in cities with population greater than 100,000 people. d. Between 23 and 30 (including 23 and 30) of them live in cities with population greater than 100,000 people.

62% of all Americans live in cities with population greater than 100,000 people. If 41 Americans are randomly selected, find the probability that a. Exactly 24 of them live in cities with population greater than 100,000 people. b. At most 27 of them live in cities with population greater than 100,000 people. c. At least 23 of them live in cities with population greater than 100,000 people. d. Between 23 and 30 (including 23 and 30) of them live in cities with population greater than 100,000 people.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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