QUESTION 26 Employees in a company experience distractions (phone calls, emails, unscheduled visits etc.) during their office hours. Given that the distraction times are right skewed with a mean of 95 minutes and a standard deviation of 18 minutes. What is the probability that for a random sample of 36 employees their mean distraction time is more than 100 minutes? 0.0004 0.0478 0.3906 0.9772

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A: Basic formulas for mean has applied.

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A: given data  μ= 3370σ = 680n = 64mean and standard error= ?

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Q: QUESTION 27 A car insurance company wanted information about the age of its customers. From a sample…

A: given data sample size (n) = 143sample mean ( x¯ ) = 37sample standard deviatio (s) = 12.898% ci for…

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A:

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A:

Q: If P(E)= 0.60, P(E or F) = 0.75, and P(E and F) = 0.10, find P(F). P(F) = (Simplify your answer.)

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A: given data σ = 0.79n = 35X¯ = 7.8293% ci for μ.

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A:

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A: Number of white balls  = 59  Number of red balls = 35 Selecting 1 from red and 5 from white balls

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A:

Q: he pmf of a discrete random variable X is as follows: X -1 0 1 2 f(x) 1/10 2/10 5/10 2/10…

A: The mean and variance of a random variable X are given by, μ= ∑XP(X)σ2=∑X2P(X)-E(X)2

Q: Employees in a company experience distractions (phone calls, emails, unscheduled visits etc.) during…

A: From the provided information, Mean (µ) = 95 Standard deviation (σ) = 18 Sample size (n) = 36

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A:

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Q: The U.S. Bureau of Prisons publishes data in Statistical Report on the times served by prisoners…

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Q: QUESTION 26 The waiting time (in minutes) of customers at a fast food outlet is normally…

A: given data, normal distribution, n=15df=n-1=15-1=14CI=0.99α=1-0.99=0.01CI is (1.83 , 2.59)

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Q: The probability that a card drawn from a pack of 52 cards will be a diamond or a king, is.

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A: Note: HI, there! As per the guidelines, we are allowed to first question when multiple questions are…