Solution to Homework 7

Problem 1
Mean: n=30, ∑x=81.5 so

=81.5/30=2.717
Median the ordered dataset is :
0.2 0.5 1.2 1.2 1.2 1.7 1.9 2.0 2.0 2.1 2.3 2.4 2.5 2.5 2.6
2.6 2.7 2.8 2.9 3.0 3.1 3.1 3.1 3.5 3.6 3.7 3.9 4.0 4.1 9.1
so the Median is (2.6+2.6)/2 = 2.6
Standard Deviation n=30, ∑x=81.5, ∑x²=291.69, so
S_xx = ∑x²-(∑x)²/n = 291.69-81.5²/30 = 70.2816
and so s = √(S_xx/(n-1)) = √(70.2816/29) = 1.56

Problem 2 Q₁=P₂₅, np/100 = 30*25/100 = 7.5 round up to 8, 8^th observation is 2.0
Q₃=P₇₅, np/100 = 30*75/100 = 22.5 round up to 23, 23^rd observation is 3.1

so we already have the 5-number summary:

Minimum Q₁ Median Q₃ Maximum

0.2 2.0 2.6 3.1 9.1

IQR = Q₃-Q₁ = 3.1-2.0 = 1.1
Boxplot:
LF = Q₁-1.5·IQR = 2.0-1.5·1.1 = 0.35 > 0.2, so left fence goes to 0.35
UF = Q₃+1.5·IQR = 3.1+1.5·1.1 = 4.75 < 9.1, so right fence goes to 4.75

Problem 3
P₁₀: np/100 = 30·10/100 = 3, third observation is 1.2
P₉₀: np/100 = 30·90/100 = 27, 27^th observation is 3.9

Problem 4
-2s = 65.6-2·12.8 = 40.0
+2s = 65.6+2·12.8 = 91.2
so 95% of the students (about 451) had a score between 40 and 91. So the others (475-451=24) are outside this interval, half with a score lower than 40, half (12) with a score more than 91. Those are the A's more than 90. Those are the A's