Given the following data set: 3 5 6 7 7 8 8 8 9 9 9 10 102 Researcher detected the technical error in the last observation and replaced 102 by 10.2. What happens to Interquartile Range (IQR) and Standard Deviation (SD)

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

fwms@lil_ben06

step-by-step explanation:

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1