Question

A manufacturer of paper used for packaging requires a minimum strength of 1600 g/cm². To check on the quality of the paper, a random sample of 10 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 160 g/cm², and the strength measurements are normally distributed.

(a) What is the approximate sampling distribution of the sample mean of n = 10 test pieces of paper?

i) The sampling distribution is nonnormal with mean μ and standard deviation 160.

ii) The sampling distribution is nonnormal with mean μ and standard deviation 160/

V10

iii) The sampling distribution is normally distributed with mean 10 and standard deviation 160.

iv) The sampling distribution is normally distributed with mean μ and standard deviation 160/

V10

v) The sampling distribution is normally distributed with mean μ and standard deviation 160.

(b) If the mean of the population of strength measurements is 1650 g/cm², what is the approximate probability that, for a random sample of n = 10 test pieces of paper, x̄ < 1600?

(Round your answer to four decimal places.)


(c) What value would you select for the mean paper strength μ in order that P(x̄ < 1600) be equal to 0.001? (Round your answer to three decimal places.)

= g/cm²

Answer 1

Sol:

a)

Option " iv " is correct.

iv). The sampling distribution is normally distributed with mean μ and standard deviation 160/

V10

b)

µ =    1600   
σ =    160                                      
n=   10                                    
                                          
X =   1600                                      
                                          
Z =   (X - µ )/(σ/√n)

= (   1600   -   1650.00   ) / (   160.000   / √   10   )

=   -0.988  
                                          
P(X ≤   1600   ) = P(Z ≤   -0.988 )

P(X ≤   1600   ) =  0.1635   

c)

Z value at    0.001   =   -3.090
Z =   (X - µ )/(σ/√n)

-3.090 = (   1600   - µ) / (   160.000   / √   10   )

µ = 1749.06

If you Satisfy with Answer, Please give me "Thumb Up". It is very useful for me.
Thank you for your support.