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?

Answer:a)  P (  1100 < X < 1400 ) = 0.755b) P (  X < 1000 ) = 0.755c) proportion ( X > 1200 ) = 65.66%d) 5.87% percentileStep-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 ) = 0.755Hence,       P (  1100 < X < 1400 ) = 0.755   ... Answerb) 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 ) = 0.0254Hence,        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 ) = 0.6566Hence,       proportion of X > 1200 = P (  X > 1200 )*100 = 65.66%   ... Answerd) 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 ) = 0.0587Hence,        Rank = proportion of X < 1050 = P (  X < 1050 )*100                  = 0.0587*100 %                    = 5.87 % ... Answer