the amount of tea leaves in a can from a particular producti

the amount of tea leaves in a can from a particular production line is normally distributed with = 110 grams and a = 25 grams. a. What is the probability that a randomly selected can will contain at least 100 grams of tea leaves? b. What is the probability that a randomly selected can will contain less than 100 grams of tea leaves? c. What is the probability that a randomly selected can will contain between I 00 and 110 grams of tea leaves? d. What is the probability that a randomly selected can will contain less than 100 grams or more than 120 grams of tea leaves? e. Approximately 83% of the cans will have at least how many grams of tea leaves?

Solution

Mean ( u ) =110
Standard Deviation ( sd )=25
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 100) = (100-110)/25
= -10/25 = -0.4
= P ( Z >-0.4) From Standard Normal Table
= 0.6554  

b)
P(X < 100) = (100-110)/25
= -10/25= -0.4
= P ( Z <-0.4) From Standard Normal Table
= 0.3446                  

c)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 100) = (100-110)/25
= -10/25 = -0.4
= P ( Z <-0.4) From Standard Normal Table
= 0.34458
P(X < 110) = (110-110)/25
= 0/25 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(100 < X < 110) = 0.5-0.34458 = 0.1554                  
              
d)
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 100) = (100-110)/25
= -10/25= -0.4
= P ( Z <-0.4) From Standard Normal Table
= 0.3446
P(X > 110) = (110-110)/25
= 0/25 = 0
= P ( Z >0) From Standard Normal Table
= 0.5
P( X < 100 OR X > 110) = 0.3446+0.5 = 0.844578