Answer:
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
