Suppose that the distribution of x = the number of items produced by an assembly

Question

Suppose that the distribution of x = the number of items produced by an assembly line during an 8-hour shift can be approximated by a normal distribution with mean value 130 and standard deviation 10. (Round your answers to four decimal places.)

(a) What is the approximate probability that the number of items produced is at most 110?
P(x 110) =   

(b) What is the approximate probability that at least 105 items are produced?
P(x 105) =   

(c) What is the approximate probability that between 115 and 139 (inclusive) items are produced?
P(115 x 139) =

Explanation / Answer

Normal Distribution
Mean ( u ) =130
Standard Deviation ( sd )=10
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
                  
a.
P(X > 110) = (110-130)/10
= -20/10 = -2
= P ( Z >-2) From Standard Normal Table
= 0.9772                  
P(X < = 110) = (1 - P(X > 110)
= 1 - 0.9772 = 0.0228                  

b.
P(X < 105) = (105-130)/10
= -25/10= -2.5
= P ( Z <-2.5) From Standard Normal Table
= 0.0062                  
P(X > = 105) = (1 - P(X < 105)
= 1 - 0.0062 = 0.9938                  

c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 115) = (115-130)/10
= -15/10 = -1.5
= P ( Z <-1.5) From Standard Normal Table
= 0.06681
P(X < 139) = (139-130)/10
= 9/10 = 0.9
= P ( Z <0.9) From Standard Normal Table
= 0.81594
P(115 < X < 139) = 0.81594-0.06681 = 0.7491                  

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