In: Statistics and Probability
The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and a standard deviation of 117 chips.
(a) Determine the 25th percentile for the number of chocolate chips in a bag.
(b) Determine the number of chocolate chips in a bag that make up the middle 9.7% of bags.
(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies?
Part a)
X ~ N ( µ = 1262 , σ = 117 )
P ( X < ? ) = 25% = 0.25
Looking for the probability 0.25 in standard normal table to
calculate critical value Z = -0.67
Z = ( X - µ ) / σ
-0.67 = ( X - 1262 ) / 117
X = 1183.61
P ( X < 1183.61 ) = 0.25
Part b)
X ~ N ( µ = 1262 , σ = 117 )
P ( a < X < b ) = 0.097
Dividing the area 0.097 in two parts we get 0.097/2 = 0.0485
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.0485
Area above the mean is b = 0.5 + 0.0485
Looking for the probability 0.4515 in standard normal table to
calculate critical value Z = -0.12
Looking for the probability 0.5485 in standard normal table to
calculate critical value Z = 0.12
Z = ( X - µ ) / σ
-0.12 = ( X - 1262 ) / 117
a = 1247.96
0.12 = ( X - 1262 ) / 117
b = 1276.04
P ( 1247.96 < X < 1276.04 ) = 0.097
Part c)
Interquartile range =Q3-Q1
Q1= 25th percentile = X = 1183.61
Q3 = 75th percentile
X ~ N ( µ = 1262 , σ = 117 )
P ( X < ? ) = 75% = 0.75
Looking for the probability 0.75 in standard normal table to
calculate critical value Z = 0.67
Z = ( X - µ ) / σ
0.67 = ( X - 1262 ) / 117
X = 1340.39
P ( X < 1340.39 ) = 0.75
So the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies = 1340.39 - 1183.61 = 156.78