Answer:
After correcting the error, IQR remains unchanged while standard deviation reduced drastically from 26.3 to 2.04
Step-by-step explanation:
Given the data:
3 5 6 7 7 8 8 8 9 9 9 10 102
Ordered data: 3, 5, 6, 7, 7, 8, 8, 8, 9, 9, 9, 10, 102
The interquartile range (IQR) for the data:
IQR = Q3 - Q1
Q3 = 3/4(n + 1) th term.
n = sample size = 13
Q3 = 3/4(14) = 10.5th term
Q3 = (10 + 11)th term / 2 = (9+9)/2 = 9
Q1 = 1/4(14) = 3.5th term
Q3 = (3 + 4)th term / 2 = (6+7)/2 = 6.5
IQR = Q3 - Q1 = 9 - 6.5 = 2.5
THE STANDARD DEVIATION :
Sqrt[Σ(X - m)²/n-1]
Using calculator :
Standard deviation = 26.3
Correcting error in the data:
3 5 6 7 7 8 8 8 9 9 9 10 10.2
Q3 = 3/4(n + 1) th term.
n = sample size = 13
Q3 = 3/4(14) = 10.5th term
Q3 = (10 + 11)th term / 2 = (9+9)/2 = 9
Q1 = 1/4(14) = 3.5th term
Q3 = (3 + 4)th term / 2 = (6+7)/2 = 6.5
IQR = Q3 - Q1 = 9 - 6.5 = 2.5
THE STANDARD DEVIATION :
Sqrt[Σ(X - m)²/n-1]
Using calculator :
Standard deviation = 2.04
After correcting the error, IQR remains unchanged while standard deviation reduced drastically from 26.3 to 2.04
