The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 12521252 and standard deviation 129129 chips. (a) what is the probability that a randomly selected bag contains between 11001100 and 14001400 chocolate chips? (b) what is the probability that a randomly selected bag contains fewer than 10001000 chocolate chips? (c) what proportion of bags contains more than 12001200 chocolate chips? (d) what is the percentile rank of a bag that contains 10501050 chocolate chips?

a)  P (  1100 < X < 1400 ) = 0.755

b) P (  X < 1000 ) = 0.755

c) proportion ( X > 1200 ) = 65.66%

d) 5.87% percentile

Step-by-step explanation:

Solution:-

- Denote a random variable X: The number of chocolate chip in an 18-ounce bag of chocolate chip cookies.

- The RV is normally distributed with the parameters mean ( u ) and standard deviation ( s ) given:

                               u = 1252

                               s = 129

- The RV ( X ) follows normal distribution:

                       X ~ Norm ( 1252 , 129^2 )  

a) what is the probability that a randomly selected bag contains between 1100 and 1400 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( x1 < X < x2 ) = P ( [ x1 - u ] / s < Z <  [ x2 - u ] / s )

- Taking the limits x1 = 1100 and x2 = 1400. The standard normal values are:

     P (  1100 < X < 1400 ) = P ( [ 1100 - 1252 ] / 129 < Z <  [ 1400 - 1252 ] / 129 )

                                        = P ( - 1.1783 < Z < 1.14728 )

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( -1.1783 < Z < 1.14728 ) = 0.755

Hence,

      P (  1100 < X < 1400 ) = 0.755   ... Answer

b) what is the probability that a randomly selected bag contains fewer than 1000 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( X < x2 ) = P ( Z <  [ x2 - u ] / s )

- Taking the limit x2 = 1000. The standard normal values are:

     P (  X < 1000 ) = P ( Z <  [ 1000 - 1252 ] / 129 )

                                        = P ( Z < -1.9535 )

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( Z < -1.9535 ) = 0.0254

Hence,

       P (  X < 1000 ) = 0.755   ... Answer

(c) what proportion of bags contains more than 1200 chocolate chips?

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( X > x1 ) = P ( Z >  [ x1 - u ] / s )

- Taking the limit x1 = 1200. The standard normal values are:

     P (  X > 1200 ) = P ( Z >  [ 1200 - 1252 ] / 129 )

                                        = P ( Z > 0.4031 )

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( Z > 0.4031 ) = 0.6566

Hence,

      proportion of X > 1200 = P (  X > 1200 )*100 = 65.66%   ... Answer

d) what is the percentile rank of a bag that contains 1050 chocolate chips?

- The percentile rank is defined by the proportion of chocolate less than the desired value.

- Compute the standard normal values for the limits of required probability using the following pmf for standard normal:

     P ( X < x2 ) = P ( Z <  [ x2 - u ] / s )

- Taking the limit x2 = 1050. The standard normal values are:

     P (  X < 1050 ) = P ( Z <  [ 1050 - 1252 ] / 129 )

                                        = P ( Z < 1.5659 )

- Use the standard normal tables to determine the required probability defined by the standard values:

       P ( Z < 1.5659 ) = 0.0587

Hence,

       Rank = proportion of X < 1050 = P (  X < 1050 )*100

                 = 0.0587*100 %  

                 = 5.87 % ... Answer

step-by-step explanation: