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The amount of nicotine in a cigarette produced by a tobacco company is a random variable with mean 2.2 mg and standard deviation 0.3 mg. Taking 100 randomly chosen cigarettes, let X; denote the nicotine content of the ith cigarette for i = 1,..., 100, and let XX; be the sample mean of the nicotine content. 100 100 i-1 a. What is the variance of X? Var(X) = Approximate the probability that X is higher than 2.25 in terms of the standard normal random variable. (The answer is given in terms of 2.25, standardized by mean and standard deviation of X.) P(X> 2.25) = P(Z >

The amount of nicotine in a cigarette produced by a tobacco company is a random variable with mean 2.2 mg and standard deviation 0.3 mg. Taking 100 randomly chosen cigarettes, let X; denote the nicotine content of the ith cigarette for i = 1,..., 100, and let XX; be the sample mean of the nicotine content. 100 100 i-1 a. What is the variance of X? Var(X) = Approximate the probability that X is higher than 2.25 in terms of the standard normal random variable. (The answer is given in terms of 2.25, standardized by mean and standard deviation of X.) P(X> 2.25) = P(Z >

MATLAB: An Introduction with Applications
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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The amount of nicotine in a cigarette produced by a tobacco company is a random variable with mean 2.2 mg and standard deviation 0.3 mg. Taking 100 randomly chosen cigarettes, let X; denote the nicotine content of the ith cigarette for i = 1,..., 100, and let X = 100 Ei=1 X; be the sample mean of the nicotine content. 100 a. What is the variance of X? Var(X) = Approximate the probability that X is higher than 2.25 in terms of the standard normal random variable. (The answer is given in terms of 2.25, standardized by mean and standard deviation of X.) P(X> 2.25) = P(Z >
Transcribed Image Text:The amount of nicotine in a cigarette produced by a tobacco company is a random variable with mean 2.2 mg and standard deviation 0.3 mg. Taking 100 randomly chosen cigarettes, let X; denote the nicotine content of the ith cigarette for i = 1,..., 100, and let X = 100 Ei=1 X; be the sample mean of the nicotine content. 100 a. What is the variance of X? Var(X) = Approximate the probability that X is higher than 2.25 in terms of the standard normal random variable. (The answer is given in terms of 2.25, standardized by mean and standard deviation of X.) P(X> 2.25) = P(Z >
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