Stat 20 Summer ’15
QUIZ 1 SOLUTIONS
Prof. Fletcher Ibser
Full Name:
SID:
Note: Show your explanations / calculations
Problem 1
Part a. Consider the following four histograms. You can assume that the data is continuous and
evenly distributed within each bar. The average of each dataset is 5 and the standard deviations
of the values in the four datasets are 0.8, 2, 2.9, and 4.7. Match each SD to its histogram. Explain
your reasoning. [3 points]
40
30
0
10
20
% per unit
30
20
0
10
% per unit
40
50
(b)
50
(a)
0
2
4
6
8
10
0
2
4
10
6
8
10
50
40
30
0
10
20
% per unit
40
30
20
0
10
% per unit
8
(d)
50
(c)
6
0
2
4
6
8
10
0
2
4
Answer: You should eyeball this using the rule of thumb that 95% of the data falls within 2 SDs
of the average. (a) has SD 2.9; (b) has SD 0.8; (c) has SD 2; (d) has SD 4.7.
Part b. For each of the four histograms, what percent of observations are greater than 7.5? Approximate to the nearest 5%. [4 points]
Answer: To estimate this percentage, add up the areas of the bars to the right of 8 and half the
area of the bar between 7 and 8. (a) has 25%; (b) has < 5%; (c) has about 15%; (d) has 45%.
Problem 2
The table below gives the typical sleep habits of five famous writers.
Name
Voltaire
Kurt Vonnegut
Maya Angelou
Victor Hugo
Charles Dickens
Sleep (hours)
4
7.5
7.5
9.5
7
Part a. Find the average and standard deviation of the number of hours of sleep. [2 points]
Answer: The average is
4 + 7.5 + 7.5 + 9.5 + 7
35.5
=
= 7.1
5
5
and the SD is
r
(0.4)2 + (2.4)2 + (−3.1)2 + (−0.1)2 + (0.4)2 √
= 3.14 ≈ 1.77
5
Part b. Whose sleep habits were more extreme, compared to the other writers – Voltaire or
Victor Hugo? [1 point]
Answer: Standardize their sleep hours:
4 − 7.1
9.5 − 7.1
≈ −1.75
versus
= 1.35
1.77
1.77
Voltaire’s typical hours of sleep are 1.75 standard units below the mean and Hugo’s are 1.35
standard units above the mean. Voltaire was more extreme.
Part c. What percent of the authors are within 1 SD of the average? [2 points]
Answer: 60%: Vonnegut, Angelou, and Dickens are within 7.1 ± 1.77 = (5.33, 8.87), while the
other two are not.
Problem 3
A jewelry store has 100 rings. The average price for a ring is $500. A thief steals the 5 most
expensive rings in the store.
Part a. The average price of the remaining 95 rings is $400. What was the average price of the 5
stolen rings? [4 points]
Answer: Let x1 , . . . , x95 be the prices of the remaining rings and x96 , . . . , x100 be the prices of the
stolen rings. We can find the sum of prices of the first 95 rings as
95
1 X
400 =
xi
95
i=1
95
X
(1)
xi = 38, 000
i=1
Use this to solve for the sum of prices of the stolen 5 rings:
100
95
X
i=1
i=1
1 X
1
500 =
xi =
100
100
50, 000 = 38, 000 +
100
X
100
X
xi +
!
xi
i=96
(2)
xi
i=96
100
X
xi = 12, 000
i=96
Finally, the average is just this sum divided by 5: $12, 000/5 = $2, 400.
Part b. The standard deviation of the prices of the 5 stolen rings is $100 and the standard deviation of the remaining 95 rings is $300. What was the standard deviation of all 100 rings? [4 points]
Answer: We’ll use the second summation formula for standard deviation to solve for the two sums
of squares:
v
u
95
u1 X
x2i − (400)2
300 = t
95
i=1
v
u 100
u1 X
100 = t
x2i − (2, 400)2
5
i=96
=⇒
95
X
x2i = 23, 750, 000
(3)
x2i = 28, 850, 000
(4)
i=1
=⇒
100
X
i=96
Now we can add together (3) and (4) to get the standard deviation of the combined list:
r
p
1
(23, 750, 000 + 28, 850, 000) − (500)2 = 276, 000 ≈ $525
SD =
100