Page 1 of 14
Stat200Introduction to Statistics Name______________________________
Answer Sheet
Record your answers and work.
Problem
Number Solution
1
(25 pts)
Answers:
(a) True
(b) True
(c) False
(d) True
(e) False
Work for (a), (b), (c), (d) and (e):
2
(5 pts)
Answer: 70%
Work:
Checkout
Time Frequency
(in minutes)
1.0 - 1.9 2
2.0 - 2.9 12
3.0 - 3.9 2
4.0 - 4.9 4
Total 20
percentage of the checkout times was less than 3 minutes = 14/20 = 0.7 = 70%
3
(5 pts)
Answer: 2nd class: 2.0 - 2.9
Work: Here, n = 20,
Median = n/2th observation when n = even,
Here, n = 20/2th obs = 10th observation. Thus median is expected to lie in 2nd class
which is 2.0 – 2.9.

Page 2 of 14
4
(5 pts)
Answer: 2.85
Work:
Checkout
Time Frequency
(in minutes) fi
mid point
(xi( fixi
1.0 - 1.9 2 1.45 2.9
2.0 - 2.9 12 2.45 29.4
3.0 - 3.9 2 3.45 6.9
4.0 - 4.9 4 4.45 17.8
Total 20 57
Mean = ∑fixi / ∑fi
= 57 / 20 = 2.85
5
(10 pts)
Answer: 0.92
Work:
Checkout
Time Frequency
(in minutes) Fi
mid point
(xi( fixi fi*xi^2
1.0 - 1.9 2 1.45 2.9 4.205
2.0 - 2.9 12 2.45 29.4 72.03
3.0 - 3.9 2 3.45 6.9 23.805
4.0 - 4.9 4 4.45 17.8 79.21
20 57 179.25
Variance = (1/n)*{ ∑ (fi*xi^2) – n*x-bar^2 }
= (1/20)*( 179.25 – 20*2.85^2)
= (1/20)*(179.25 – 162.45)
= (1/20)*(16.8)
= 0.84
Standard deviation = sqrt(variance)
= sqrt(0.84)
= 0.9165
= 0.92 (approximately)

Page 4 of 14
9
(10 pts)
Answer: 1/221
Work: Probability (1st card is ace and 2nd card is also an ace) = ?
Probability of drawing 1st card as ace = 4/52
Probability of drawing 2nd card as ace without replacement = 3/51
Thus required probability = (4/52)*(3/51) = (1/13)*(1/17) = 1/221
10
(10 pts)
Answer: 1/169
Work: Probability (1st card is ace and 2nd card is also an ace) = ?
Probability of drawing 1st card as ace = 4/52
Probability of drawing 2nd card as ace with replacement = 4/52
Thus required probability = (4/52)*(4/52) = (1/13)*(1/13) = 1/169
11
(5 pts)
Answer:yes, events are independent in case of with replacement because P(A) * P(B)
= P(A and B)
Work: two events are said to be independent if P(AandB) = P(A) * P(B)
In case of with replacement,
LetA denotethe event of drawing 1st ace and event B denote the event of drawing 2nd
ace.
Now, P(A) = Probability of drawing 1st card as ace = 4/52 = 1/13
P(B) = Probability of drawing 2nd card as ace (with replacement)= 4/52 = 1/13
P(A) * P(B) = 1/13 * 1/13 = 1/169
P(A and B) = P(drawing 1st card as ace and 2nd card as ace without replacement) =
(4/52 * 4/52) = 1/169
Since, P(A and B) = P(A)*P(B), thus events are independent
12
(10 pts)
Answer: 0.25
Work:
Let event A denote STAT200
And event B denote PHY300
Here total number of juniors in college is 1000

Stat200 Introduction to Statistics

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