Customers arrive at a movie theater at the advertised movie time only to find that they haveto sit through several previews and prepreview ads before the movie starts. Many complainthat the time devoted to previews is too long (the Wall Street Journal, October 12, 2012).A preliminary sample conducted by the Wall Street Journal showed that the standarddeviation of the amount of time devoted to previews was 4 minutes. Use that as a planningvalue for the standard deviation in answering the following questions.a. If we want to estimate the population mean time for previews at movie theaters with amargin of error of 75 seconds, what sample size should be used? Assume 95% confidence

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A: From the given information,Mean (µ) =600Standard deviation (σ) = 100Let the test score be a x which…

Q: O 0.0226 O 0.4885 O 0.5115 O 0.15 O 0.0113

A: Sample mean, Standard deviation, s= 5degrees of freedom, df= 80-1=79The null and alternative…

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A: Given Data: Let X represent the random variable Mean=9 Standard deviation=1.1 X~ Normal (9, 1.12)…

Q: Suppose that the quarterly sales levels among health care information systems companies are…

A: Given data,Mean μ=11sd      σ=1.2P(X<x)=20%x=?

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A: The provided data is,

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A: given data normal distribution μ = 12σ = 1.2

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A: Solution-: Given:μ=40,σ=11.2 We want to find  the proportion of customers who use ATMs between 30…

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A:

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A: Given: Mean= μ== 67.5 Standard deviation= σ = 12.5

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A:

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A: here given, mean = 50 minute standard deviation  = 6 minute

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A: mean=8 standard deviation=1.1

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A:

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A:

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Q: Suppose that the quarterly sales levels among health care information systems companies are…

A: Given Information: Suppose that the quarterly sales levels among health care information systems…

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Q: Find an estimator for b when a = 0 using the method of moments.

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A: From the provided information,

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Q: Suppose that the quarterly sales levels among health care information systems companies are…

A: Given information- Population mean µ = 8 million Standard deviation σ = 1.1 million Z = -1.2816…

Q: Suppose that the quarterly sales levels among health care information systems companies are…

A: The objective of this question is to find the sales level that separates the bottom 15% of all…

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A:

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A: given datanormal distributionμ = 528n = 110top 2.5% of student get scholarship

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A: Here, σ=45, confidence level=90%, E=5.

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A: Answer  Mean =12 standard deviation =1.3

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A: Introduction Each possible situation of a random phenomenon is indicated by the probability…