Question

Using R, answer the following question. The accompanying data set contains quiz scores for 12 students in a chemistry class: 18, 15, 5, 8, 15, 20, 2, 16, 10, 12, 20, 15.

Find a measure of central tendency that separates the data into two groups such that each group consists of 50% of the scores above and 50% of the scores below that measure.
Find a measure of central tendency that represents the average of the 12 quiz scores.
Find the standard deviation and variance for this dataset.

Answer 1

• Median = 15
• Mean = 13
• Standard Deviation = 5.86 ; Variance = 34.33

Explanation:

1. Measure of central tendency which divides data into 2 equal halves (50% below it , 50% after it is Median

Median Calculation :

Data arranged : 2 , 5 , 8 , 10 , 12 , 15 , 15 , 15 , 16 , 18 , 20 , 20

N = 12 (Even)

Median = (N / 2)th + (N/2 + 1)th observations

2

= [ ( 12/2 )th obs + (12/2 + 1)th obs ] / 2

= [6th + 7th ] / 2

= [15 + 15] / 2

= 15

2. Mean Calculation

Mean = Σ X / N

Σ X = 18 + 15 + 5 + 8 + 15 + 20 + 2 + 16 + 10 + 12 + 20 +15 = 156

N = 12

Mean [ U ] = 156 / 12

Mean = 13

3. Standard Deviation [σ] & Variance Calculation :

Formula : Variance = σ^2 = [ Σ (X - U )^2 ] / N

X - U → 2 , 5 , 8 , 10 , 12 , 15 ,15 , 15 , 16 , 18 , 20 , 20

X - (15) → -13,-10,-7,-5 , -3 , 0 , 0 , 0 , 1 , 3 , 5 , 5

(X - U)^2 →169,100,49,25 , 9 , 0 , 0 , 0 , 1 , 9 , 25 , 25 [ Σ = 412]

Variance = 412 / 12 = 34.33

Standard Deviation = √ Var

√ 34.33 = 5.86