Question

A manufacturer of paper used for packaging requires a minimum strength of 1800 g/cm². To check on the quality of the paper, a random sample of 13 pieces of paper is selected each hour from the previous hour's production and a strength measurement is recorded for each. The standard deviation σ of the strength measurements, computed by pooling the sum of squares of deviations of many samples, is known to equal 180 g/cm², and the strength measurements are normally distributed.

a) If the mean of the population of strength measurements is 1850 g/cm², what is the approximate probability that, for a random sample of n = 13 test pieces of paper,

x < 1800? (Round your answer to four decimal places.)

b) What value would you select for the mean paper strength μ in order that P(x < 1800) be equal to 0.001? (Round your answer to three decimal places.)

Answer 1

Given: 6:180 TP.o a) bere el=1850, n=13 We have to find PCXC 1800) P ( x< 1800) P(x-4 < 1800 -7850 180 - PONCO,1) < .0.2278) : 0.3906 PCX<1800 ) 6) We have to find 'll such that PC X< 180 o ) 0.0 o | (^-4 ( 18 0 0 - 4 1) 0.00! - 180 1800 - 4 180 ztc0.001) 1800-4 180 -3.09 2170 el : 25356.2