Lies and Statistics
An independent survey conducted in a large school indicated the lying habits of the students were as follows
category / percentage of students / percentage of lies told of all statements made): (A) Truthful students / 10% / 0%;
(B) Students who seldom lied / 20% / 20%;
(C) Students who evenly lied / 40% / 50%;
(D) Students who frequently lied / 20% / 80%;
(E) Students who were liars / 10% / 100%.
The same group of students were given a multiple choice questionnaire and were asked to indicate the category (A, B, C, D, or E) which they belonged to. Obviously students would indicate their category, depending upon their nature – eg: All truthful students would put themselves in category A, but Liars would never indicate category E.
The problem is, what would be the percentages of students in various categories as self indicated by the students? Further if a second self-assessment survey is conducted asking the students whether they had truthfully answered the first self assessment survey having only two categories (T) Told the truth, and (F) Lied, what would the results tell in percentage terms?
category / percentage of students / percentage of lies told of all statements made): (A) Truthful students / 10% / 0%;
(B) Students who seldom lied / 20% / 20%;
(C) Students who evenly lied / 40% / 50%;
(D) Students who frequently lied / 20% / 80%;
(E) Students who were liars / 10% / 100%.
The same group of students were given a multiple choice questionnaire and were asked to indicate the category (A, B, C, D, or E) which they belonged to. Obviously students would indicate their category, depending upon their nature – eg: All truthful students would put themselves in category A, but Liars would never indicate category E.
The problem is, what would be the percentages of students in various categories as self indicated by the students? Further if a second self-assessment survey is conducted asking the students whether they had truthfully answered the first self assessment survey having only two categories (T) Told the truth, and (F) Lied, what would the results tell in percentage terms?
posted by Rajesh Lal at
7:04 AM
6 Comments:
Before I attempt to solve this... you have a total of 110% students... plz to be fixing this.
It sounds like a story from FOX news. The reporter is making it all up.
T) Told the truth =55%, (F) Lied =55%
110 % of students
Hey Cry Wolf
you are right, updated the question
When I did this I came up with:
A:22.5%
B:27.5%
C:27.5%
D:12.5%
E:10%
If it was only True or False, it would be 50% true and 50% False.
Assume, since the school is a "large" school, and assume each survey is sufficiently large, that the numbers will approach their average values.
***1st survey,
What percentage of students whould indicate themselves in each of the categories ?
Assume that students who lie, will randomly select one of the other 4 categories.
Category_T= Sum_category
Category_T= Honest + Sum(Liars)/4
i.e. those who select the category consist of those who are actually in it and tell the truth and of all the liars who aren't in it but say they are.
Category_H= Honest
Category_L= Liars/4
A_H= 10*100%=10
A_L= 10*0%/4=0
B_H= 20*80%=16
B_L= 20*20%/4= 1
C_H= 40*50%= 20
C_L= 40*50%/4= 5
D_H= 20*20%= 4
D_L= 20*80%/4 = 4
E_H= 10*0%=0
E_L= 10*100%/4= 2.5
A_T= A_H + B_L +C_L +D_L +E_L
A_T= 10+ 1 + 5 + 4 + 2.5
A_T= 22.5
Similiarly
B_T= A_L + B_H +C_L +D_L + E_L
B_T= 0 + 16 + 5 + 4 + 2.5
B_T= 27.5
C_T, D_T, E_T solved in same manner.
ANSWER : PART1
A: 22.5
B: 27.5
C: 27.5
D: 12.5
E: 10.0
***2nd Survey
Those who will claim to have told the truth in the first survey consists of:
1) Those who did tell the truth and are telling the truth again i.e. 1*1= 1
2) Those who lied and are lieing again i.e. -1*-1= 1
Those who claim to have lied consists of :
1) Those who did tell the truth and now lying i.e. 1*-1= -1
2) Those who lied and are now telling the truth i.e. -1*1= -1
T= Double honest + Double liars
T_A= 10*100%*100% + 10*0%*0%=10
T_B= 20*80%*80% + 20*20%*20%=13.6
T_C= 40*50%*50% + 40*50%*50%= 20
T_D= 20*20%*20%+20*80%*80% = 13.6
T_E= 10*0%*0% + 10*100%*100% = 10
T= T_A +T_B +T_C +T_D +T_E
T= 67.2
F=1-T= 32.8
F should be 1-T but as a cross check we can calculate F
F= Truth the lie + Lie then truth
F_A= 10*100%*0% + 10*0%*100%= 0
F_B= 20*80%*20% + 20*20%*80%= 6.4
F_C= 40*50%*50% + 40*50%*50%= 20
F_D= 20*20%*80%+20*80%*20% = 6.4
F_E= 10*0%*100% + 10*100%*0% = 0
F= F_A +F_B +F_C +F_D +F_E
F= 32.8
Result checks out.
ANSWER: PART2
T=67.2
F=32.8
Interesting result:
Students portray themselves as more honest than they really are.
Cam
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