Question

Suppose that the distribution of the number of items x produced by an assembly line during an 8-hr 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 probability that the number of items produced is at most 110?
P(x ? 110) =  

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

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

Answer 1

Solution :

Given that ,

mean = = 130

standard deviation = = 10

(a)

P(x 110) = P((x - ) / (110 - 130) / 10) = P(z -2)

Using standard normal table,

P(x 110) = 0.0228

Probability = 0.0228

(b)

P(x 105) = 1 - P(x < 105)

= 1 - P((x - ) / (105 - 130) / 10)

= 1 - P(z -2.5)  

= 1 - 0.0062

= 0.9938

P(x 15) = 0.9938

Probability = 0.9938

(c)

P(115 x 139) = P((115 - 130 / 10) (x - ) / (139 - 130) / 10) )

P(13.7 x 28.9)  = P(-1.5 z 0.9)

P(13.7 x 28.9) = P(z 0.9) - P(z -1.5)

P(13.7 x 28.9) = 0.8159 - 0.668 = 0.7491

Probability = 0.7491