The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.974 g and a standard deviation of 0.284 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 34 cigarettes with a mean nicotine amount of 0.896 g. Assuming that the given mean and standard deviation have NOT changed, find the probability of randomly seleting 34 cigarettes with a mean of 0.896 g or less. P(M < 0.896 g) =  Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

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A: Given,sample size(n)=10mean(μ)=2.53standard deviation(σ)=0.04

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Q: A test score in the top 2% is required. The test is designed to have a mean of 100 and a std dev. of…

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A:

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A:

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.