Please show work if possible. The true length of boards cut at a mill with a lis
ID: 2950301 • Letter: P
Question
Please show work if possible.The true length of boards cut at a mill with a listed length of 10feet is normally distributed with a mean of inches and astandard deviation of 120 inch. What is theprobability that the sample mean for a randomly selected sample of32 boards will be less than121.015 inches?
An international high-tech company wants to estimate the meannumber of years of college education completed by its employees. Asample of 56 employees yields a sample mean of2.00 years. Assume that the population standarddeviation for the number of years of college completed =0.800 years. Using a 95% level ofconfidence, find the margin of error in estimating a confidenceinterval for the true population mean number of years of collegeeducation completed.
Explanation / Answer
I think you may have written down the wrong standard deviationbecause it is seems way too big, but here it is anyways.U=10feet=120inches standard deviation(s)=120inches
trying to find P(xbar < 121.015)
the zscore for xbar =(xbar-U)/(s/squareroot of n) = .05
so P(z<.05) =.5199
b)
trying to find E which equals SE xZ*
SE= population standard deviationdivided by the square root of the sample size Z* = the z-score at an area of(1-.95)/2 (i.e .025), which turns out to be 1.960
so E = 1.960 x .2673 =.5238
U=10feet=120inches standard deviation(s)=120inches
trying to find P(xbar < 121.015)
the zscore for xbar =(xbar-U)/(s/squareroot of n) = .05
so P(z<.05) =.5199
b)
trying to find E which equals SE xZ*
SE= population standard deviationdivided by the square root of the sample size Z* = the z-score at an area of(1-.95)/2 (i.e .025), which turns out to be 1.960
so E = 1.960 x .2673 =.5238