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30. for the following population of n = 6 scores: 2, 9, 6, 8, 9, 8 a. c…

Question

  1. for the following population of n = 6 scores: 2, 9, 6, 8, 9, 8

a. calculate the range and the standard deviation. (use either definition for the range—see page 113.)

b. add 2 points to each score and compute the range and standard deviation again.

c. describe how adding a constant to each score influences measures of variability.

Explanation:

Step1: Calculate range for part a

Range = Max - Min
The scores are 2, 9, 6, 8, 9, 8. Max = 9, Min = 2.
Range = 9 - 2 = 7

Step2: Calculate mean for part a

$\bar{x}=\frac{\sum_{i = 1}^{N}x_{i}}{N}$
$\sum_{i=1}^{6}x_{i}=2 + 9+6 + 8+9 + 8=42$
$N = 6$
$\bar{x}=\frac{42}{6}=7$

Step3: Calculate squared - deviations for part a

$(x_1-\bar{x})^2=(2 - 7)^2=25$
$(x_2-\bar{x})^2=(9 - 7)^2 = 4$
$(x_3-\bar{x})^2=(6 - 7)^2=1$
$(x_4-\bar{x})^2=(8 - 7)^2 = 1$
$(x_5-\bar{x})^2=(9 - 7)^2=4$
$(x_6-\bar{x})^2=(8 - 7)^2=1$
$\sum_{i = 1}^{6}(x_{i}-\bar{x})^2=25 + 4+1+1+4+1=36$

Step4: Calculate standard deviation for part a

$S=\sqrt{\frac{\sum_{i = 1}^{N}(x_{i}-\bar{x})^2}{N}}$
$S=\sqrt{\frac{36}{6}}=\sqrt{6}\approx2.45$

Step5: Add 2 to each score for part b

The new scores are 4, 11, 8, 10, 11, 10.
Max = 11, Min = 4
Range = 11 - 4 = 7

Step6: Calculate new mean for part b

$\sum_{i=1}^{6}x_{i}=4 + 11+8 + 10+11 + 10=54$
$\bar{x}=\frac{54}{6}=9$

Step7: Calculate new squared - deviations for part b

$(x_1-\bar{x})^2=(4 - 9)^2=25$
$(x_2-\bar{x})^2=(11 - 9)^2 = 4$
$(x_3-\bar{x})^2=(8 - 9)^2=1$
$(x_4-\bar{x})^2=(10 - 9)^2 = 1$
$(x_5-\bar{x})^2=(11 - 9)^2=4$
$(x_6-\bar{x})^2=(10 - 9)^2=1$
$\sum_{i = 1}^{6}(x_{i}-\bar{x})^2=25 + 4+1+1+4+1=36$

Step8: Calculate new standard deviation for part b

$S=\sqrt{\frac{\sum_{i = 1}^{N}(x_{i}-\bar{x})^2}{N}}$
$S=\sqrt{\frac{36}{6}}=\sqrt{6}\approx2.45$

Step9: Describe the effect for part c

Adding a constant to each score does not change the range or the standard deviation. The spread of the data relative to each other remains the same.

Answer:

a. Range = 7, Standard deviation $\approx2.45$
b. Range = 7, Standard deviation $\approx2.45$
c. Adding a constant to each score does not affect measures of variability (range and standard - deviation remain the same).