The weight of a mechanical part is known to be normally distributed with mean of

Question

The weight of a mechanical part is known to be normally distributed with mean of 100 grams and variance of 9 grams.

a.) What is the probability that a randomly selected part has a weight less than 102 grams?

b.) what is probability that a randomly selected part has a weight between 99 grams and 103 grams?

c.) What is the probability that the sample mean of 16 of these parts has a weight less than 102?

d.) What is the probability that the sample mean of 16 of these parts has a weight between 99 grams and 103 grams?

Explanation / Answer

Mean ( u ) =100
Standard Deviation ( sd )=9
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
              
a.
P(X < 102) = (102-100)/9
= 2/9= 0.2222
= P ( Z <0.2222) From Standard Normal Table
= 0.5879                  

b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 99) = (99-100)/9
= -1/9 = -0.1111
= P ( Z <-0.1111) From Standard Normal Table
= 0.45576
P(X < 103) = (103-100)/9
= 3/9 = 0.3333
= P ( Z <0.3333) From Standard Normal Table
= 0.63056
P(99 < X < 103) = 0.63056-0.45576 = 0.1748                  

c.
P(X < 102) = (102-100)/9/ Sqrt ( 16 )
= 2/2.25= 0.8889
= P ( Z <0.8889) From Standard NOrmal Table
= 0.813                  

d.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 99) = (99-100)/9/ Sqrt ( 16 )
= -1/2.25
= -0.4444
= P ( Z <-0.4444) From Standard Normal Table
= 0.32836
P(X < 103) = (103-100)/9/ Sqrt ( 16 )
= 3/2.25 = 1.3333
= P ( Z <1.3333) From Standard Normal Table
= 0.90879
P(99 < X < 103) = 0.90879-0.32836 = 0.5804

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