Question

You work at a manufacturing plant and want to improve the average time needed to create and package a rocking chair. Based on previous information you know the time needed to create and package the chair is normally distributed with a mean of 13.5 hours and a standard deviation of 2.75 hours.

a) What percentage of chairs would take less than 12 hours to complete? What percentage of chairs would take more than 15 hours to complete? If a chair takes 9 hours or less to complete is this unusual? Justify this answer

b) You take a sample of 50 times needed to create and package this chair and find the average is 13 hours. Would it be unusual for you to find an average time less than or equal to 13 hours? Explain fully

c) You would like to offer a bonus if the 50 workers on a shift would adopt a new process improve the average production time. You want this average to be at the tenth percentile. What amount of time is the tenth percentile?

Answer 1

a)

µ =    13.5      
σ =    2.75      
          
P( X ≤    12   ) = P( (X-µ)/σ ≤ (12-13.5) /2.75)  
=P(Z ≤   -0.545   ) =   0.2927

µ =    13.5              
σ =    2.75              
                  
P ( X ≥   15.00   ) = P( (X-µ)/σ ≥ (15-13.5) / 2.75)          
= P(Z ≥   0.545   ) = P( Z <   -0.545   ) =    0.2927

µ =    13.5      
σ =    2.75      
          
P( X ≤    9   ) = P( (X-µ)/σ ≤ (9-13.5) /2.75)  
=P(Z ≤   -1.636   ) =   0.0509

Unusual as probability is just 5.09%

b)

µ =    13.5      
σ = 2.75 / sqrt(50) =  0.38890873      
          
P( X ≤    13   ) = P( (X-µ)/σ ≤ (13-13.5) /0.388908729652601)  
=P(Z ≤   -1.286   ) =   0.0993

Unusual becasue probability is less than 10%

c)

µ=   13.5                  
σ =    0.38890873                  
proportion=   0.1                  
                      
Z value at    0.1   =   -1.28   (excel formula =NORMSINV(   0.1   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -1.28   *   0.38890873   +   13.5  
X   =   13.00   (answer)

THANKS

revert back for doubt

please upvote