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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)

Sagot :

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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

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