\"Durable press\" cotton fabrics are treated to improve their recovery from wrin

Question

"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. The breaking strength of untreated fabric is normally distributed with mean 51.6 pounds and standard deviation 2.9 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 20.1 pounds and standard deviation 1.6 pounds. A clothing manufacturer tests 5 specimens of each fabric. All 10 strength measurements are independent. (Round your answers to four decimal places.)

a) what is the probability that the mean breaking strength of the 5 untreated specimens exceeds 50 pounds.

b) what is the probability that the mean breaking strength of the 5 untreated specimens is at least 25 pounds greater than the mean strength of the 5 untreated specimens

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    50      
u = mean =    51.6      
n = sample size =    5      
s = standard deviation =    2.9      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.233692677      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.233692677   ) =    0.891341278 [ANSWER]

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b)

Calculating the standard deviations of each group,              
              
s1 =    2.9          
s2 =    1.6          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    5          
n2 = sample size of group 2 =    5          

Hence,

sD =    1.481215717          
              
Also, the mean difference is

u = (51.6-20.1) = 31.5

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    25      
u = mean =    31.5      
          
s = standard deviation =    1.481215717      
          
Thus,          
          
z = (x - u) / s =    -4.388287219      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -4.388287219   ) =    0.999994288 [ANSWER]

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