SOLUTION: I am having so much difficulty figuring these out.
Suppose the replacement times for washing Machines are normally distributed with a mean of 9.3 years and a standard deviation of
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Suppose the replacement times for washing Machines are normally distributed with a mean of 9.3 years and a standard deviation of
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Question 795308: I am having so much difficulty figuring these out.
Suppose the replacement times for washing Machines are normally distributed with a mean of 9.3 years and a standard deviation of 1.1 years. Find the probability that 70 randomly selected washing Machines will have a mean replacement time less than 9.1 years. Found 2 solutions by rothauserc, stanbon:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! We are given a normal distribution and 70 randomly selected washing machines
When we are given a randomly selected population, the mean applies but we need to recalculate the standard deviation for the sample by dividing the population standard deviation by the square root of the sample size, so here we go
mean is 9.3 and standard deviation = 1.1 / square root (70) = .13
we are asked to determine the probability(X< 9.1), first we need to calculate the associated z-value
z-value = 9.1 - 9.3 / .13 = -1.5
now consult the negative z table and we see that
Pr(X < 9.1) = .0668 = .07 = 7%
You can put this solution on YOUR website! Suppose the replacement times for washing Machines are normally distributed with a mean of 9.3 years and a standard deviation of 1.1 years. Find the probability that 70 randomly selected washing Machines will have a mean replacement time less than 9.1 years.
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z(9.1) = (9.1-9.3)/(1.1/sqrt(70)) = -0.2/(0.1315) = -1.5209
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P(x-bar<9.1) = P(z<-1.5209) = normalcdf(-100,-1.5209)= 0.0641
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Cheers,
Stan H.
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