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 28th percentile for the number of chocolate chips in a bag. (Round to the nearest whole number as needed.)(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.
(Round to the nearest whole number as needed. Use ascending order.)
(c) What is the interquartile range of the number of chocolate
chips in a bag of chocolate chip cookies? Also, can you please
tell me how I can use statcrunch or the TI-84 calculator to get
each answer.
Problem statement:- Number of chocolate chips in a bag of chocolate chip cookies is considered to be distributed normally. Based on this information we have to answer few questions.
Given:-Number of chocolate chips in a bag of chocolate chip cookies is considered to be distributed normally with mean 1262 chips and standard deviation of 117 chips. Based on this information we have to answer few questions.
Solution:-
Problem (a):- We have to determine 28th percentile for the number of chocolate chips in a bag.
We have to compute the certain value for number of chocolate chips below which there are 28% of chocolate chips in the cookies.
=Probability ( Xx)=0.28.
Using the normal distribution population parameters , mean=1262 and standard deviation =117, in the formula,
P(Z=((x-1262)/117))=0.28.
Solving for x we get x=1193.8, approximately 1194.
Approach in TI-84 calculator
In TI-84 calculator go to DISTR (2nd VARS).
Select option 3: invNorm
Use the following syntax :-invNorm(0.28,1262117)
This should return the answer as x=1193.8, approximately 1194.
Problem (b):- We have to compute the number of chocolate chips in a bag that make up the middle 95% of bags.
Similar to the problem (a), we have to find number of chocolate chips in a bag which are at 97.5 percentile , 2.5 percentile and take a range to obtain middle 95% of bags.
=Probability ( Xx2)=0.975, Probability ( Xx1)=0.0.025.
Using the normal distribution population parameters , mean=1262 and standard deviation =117, in the formula,
P(Z=((x2-1262)/117))=0.975 and.P(Z=((x1-1262)/117))=0.97
x1=1032.7,x2=1491.3.
The range between x1 and x2 is approximately 1033 to 1491 which constitutes middle 95% of the bags. Difference between them is 1491-1033=458.
Approach in TI-84 calculator
In TI-84 calculator go to DISTR (2nd VARS).
Select option 3: invNorm
Use the following syntax :-invNorm(0.975,1262117) , this computes x2
Use the following syntax :-invNorm(0.025,1262117) , this computes x1
Then find the range and difference between x1 and x2 as required.
Problem (c):- Find Inter quartile range (IQR)
IQR is defined as the difference between third quartile (Q3) and first quartile (Q1) Third quartile is basically 75 percentile and Q1 is basically 25 percentile.
Using the similar approach as in problem (a) and (b),we get,
To compute75 percentile-Q3=P(Z=((Q3-1262)/117))=0.75
To compute25 percentile-Q1=P(Z=((Q1-1262)/117))=0.25
Solving for Q3 and Q1 we get Q3=1341 and Q1=1183
IQR= Q3-Q1=1341-1183=158
Approach in TI-84 calculator
In TI-84 calculator go to DISTR (2nd VARS).
Select option 3: invNorm
Use the following syntax :-invNorm(0.75,1262117) , this computes Q3
Use the following syntax :-invNorm(0.25,1262117) , this computesQ1
Then find difference between Q3 and Q1 as required to compute IQR.