The number of chocolate chips in an 18ounce bag of chocolate

The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 1260 and standard deviation 129 chips. (a) What is the probability that a randomly selected bag contains between 1000 and 1400 chocolate chips? (b) What is the probability that a randomly selected bag contains fewer than 1050 chocolate chips? (c) What proportion of bags contains more than 1200 chocolate chips? (d) What is the percentile rank of a bag that contains 1425

Solution

Normal Distribution
Mean ( u ) =1260
Standard Deviation ( sd )=129
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 1000) = (1000-1260)/129
= -260/129 = -2.0155
= P ( Z <-2.0155) From Standard Normal Table
= 0.02193
P(X < 1400) = (1400-1260)/129
= 140/129 = 1.0853
= P ( Z <1.0853) From Standard Normal Table
= 0.8611
P(1000 < X < 1400) = 0.8611-0.02193 = 0.8392                  
b)
P(X < 1050) = (1050-1260)/129
= -210/129= -1.6279
= P ( Z <-1.6279) From Standard Normal Table
= 0.0518                  
c)
P(X > 1200) = (1200-1260)/129
= -60/129 = -0.4651
= P ( Z >-0.465) From Standard Normal Table
= 0.6791                  
d)
   P(X < 1425) = (1425-1260)/129
= 165/129= 1.2791
= P ( Z <1.2791) From Standard Normal Table
= 0.8996                  
= 89.96% it will be almost 90% percentile