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1.A drill make holes whose diameter X are Normaldistribution. (a) If it is known

ID: 2951334 • Letter: 1

Question

1.A drill make holes whose diameter X are Normaldistribution. (a) If it is known that the middle 95% of the holes drilledhave a diameter b/w 2.57mm and 3.25mm, find the mean and standarddeviation of the diameter of the holes. (b) Interpret the inequanlity X>3 in the context of thisproblem and find its proportion. Interpret such proportion in thecontext of the problem. (c) find the diameter of the widest 32% of the holes. (d) find the diameter of the middle 32% of the holes. (e) find the narrowest a drilled hole can be in order to beamong the 32%of the holes with a diameter smaller than 3mm. p/s help me show working on all the answer. 1.A drill make holes whose diameter X are Normaldistribution. (a) If it is known that the middle 95% of the holes drilledhave a diameter b/w 2.57mm and 3.25mm, find the mean and standarddeviation of the diameter of the holes. (b) Interpret the inequanlity X>3 in the context of thisproblem and find its proportion. Interpret such proportion in thecontext of the problem. (c) find the diameter of the widest 32% of the holes. (d) find the diameter of the middle 32% of the holes. (e) find the narrowest a drilled hole can be in order to beamong the 32%of the holes with a diameter smaller than 3mm. p/s help me show working on all the answer.

Explanation / Answer

a) we know that z-scores of -2 to 2 will give a probability ofabout 95% normalcdf(-2,2) = 0.9545 z=(x-)/ x- = z (2.57 - )= -2 (3.25 - )= 2 (2.57 -) +(3.25 - ) = 0 2 = 5.82 = 2.91 = (x-)/z = (2.57 - 2.91)/-2 = 0.17 b)x>3 means diameter greater than 3mm. it's proportion isP(x>3) = P(z>3) z = (x-)/ = (3-2.91)/0.17 = 0.529 P(z> 0.529) = normalcdf(0.529, 100) = 0.2984 29.84% of the holes havea diameter greater than 3mm c) zof 0.32 = invNorm(0.32) = -0.4676 x = z+ = 2.83mm d) (1-0.32)/2 = 0.34 zof 0.34 = invNorm(0.34) = -0.4125 x = z+ = 2.83988mm and (0.4125)(0.17)+(2.91) =2.980mm e) P(y<x<3) = 0.32 P(y<z<0.529) = 0.32