Download Midterm 1 Solutions

Name:
Student Number:
STAT 270 Surrey: Spring Semester 2015
Midterm Examination # 1 Solutions
22 January 2015
Instructor: Richard Lockhart
Instructions: This is a closed book test but you are allowed one sheet of US letter sized
paper with as much on it as you want. You may use a calculator but not a computer or
a phone. Your work will be marked for clarity of explanation. Explain what assumptions
you are making and comment if those assumptions seem unreasonable. I want the answers
written on the paper. The exam is out of 12.
1. The daily maximum temperatures for January 13, 14 and 15 were 5.2, 4.4 and 8.8
degrees Celsius. What were the mean, median and standard deviation of the daily
maximum temperatures for these three days last week? [2 marks]
The sorted list is 4.4, 5.2, 8.8 so the middle number is 5.2. The median is 5.2 degrees.
We find
X
X
xi = 18.4 and
x2i = 123.84
so
x̄ =
and
18.4
= 6.13 degrees
3
r
123.84 − (18.4)2 /3 √
= 5.493 = 2.34 degrees.
2
Half a mark each for median and mean. One mark for the SD. Minus 0.5 for forgetting
the units. I won’t be fussy about how you describe the units or about rounding off.
Some people forgot to take a square root. Some people had units for the mean and
median and no units for the SD. Some people didn’t sort the list before taking the
middle value when finding the median.
s=
2. You could convert those temperatures to the Fahrenheit scale by the formula
Fahrenheit = 1.8 × Celsius + 32
What would the mean, median and standard deviation have been for those 3 days if
we had measured in Fahrenheit? [1 mark]
Letting F denote Fahrenheit and C denote Celsius, we have
F̃ = 1.8C̃ + 32 = 1.8 × 5.2 + 32 = 41.36 degrees
1
and
The SD is
F̄ = 1.8C̄ + 32 = 1.8 × 6.13 + 32 = 43.04 degrees.
sF = 1.8sC = 1.8 × 2.34 = 4.22 degrees.
Some people did all the arithmetic again after converting each temperature to Fahrenheit, which is not wrong so they got the marks but it was a lot of extra work.
3. If I had used one day in January, one day last September and one day last July instead
of 3 days last week would the standard deviation have been higher or lower? Explain.
[1 mark]
The temperatures in different seasons would be more different from each other so the
SD would be higher since it measures spread. Right answer is 0.5 and a truly coherent
explanation is required for the other 0.5. I felt that some people believed that a higher
average implied a higher SD but that would be wrong; it is the spread of the numbers
that matters.
4. I have two 4 sided dice – one red, one green. Each has the 4 sides labelled 1, 2, 3, and
4. When I toss one of these dice one side ends up face down and the number on the
face down side is the one I have tossed. If I toss the two dice what is the sample space?
[1 mark]
You could do well to arrange the list in the form of a 4 by 4 table. I list them here
with the result on the red die first: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4),
(3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4). You need coherent clear notation and
16 outcomes to get the mark. Some people said “S = 16” showing that they knew how
many elements were in S but not providing a list of the elements in any way. There is
a precise meaning to an equal sign and I am fussy about it.
5. In the previous problem what is the event that the sum of the two numbers tossed is
4? [1 mark]
The event is {(1, 3), (2, 2), (3, 1)} in the notation I used in the previous problem. Yours
has to be clear and related to your answer to the previous problem. I took off half a
mark if you told me P (A) = 3/16 because you didn’t make clear what the anwer to
the question I actually asked was. I did not ask for P (A); I asked what A was.
6. Let A be the event that the sum of the two numbers tossed is 4 and B be the event
that the sum of the two numbers tossed is even (one of 2, 4, 6, or 8, that is). Find
P (A|B) and P (B|A). [2 marks]
The event A I listed above; it has P (A) = 3/16 because A has 3 elements and the
sample space has 16 and all 16 are equally likely. The event B is
B = {(1, 1), (1, 3), (2, 2), (3, 1), (2, 4), (3, 3), (4, 2), (4, 4)}
which has 8 elements so P (B) = 8/16 = 1/2. The event AB is just A so
P (B|A) =
P (A)
P (AB)
=
=1
P (A)
P (A)
2
and
P (A|B) =
P (A)
3/16
3
P (AB)
=
=
= = 0.375.
P (B)
P (B)
8/16
8
The answers are 0.5 marks each; your explanation is the other mark and it needs
to be clear. Some people computed P (AB) = P (A)P (B), assuming A and B are
independent which is wrong. Some people seemed to think that P (AB) might be
different from P (BA) – not true because AB = BA.
7. Consider a province in which all license plates start with 3 letters and then have 3
numbers. Suppose the license plate I get is picked at random from the set of all such
license plates. What is the chance I get RAL 007? [2 marks]
There are 26 choices for each letter and 10 for each number so there are 263 × 103 =
17, 576, 000 possible license plates with the required pattern. Only 1 of these matches
my plate so the chance is
1
= 5.69 × 10−8 .
17, 576, 000
There are many ways to do this problem and I marked them all right but there were
also many ways to get it completely wrong. Order matters so binomial coefficients
don’t enter in to the problem. Some people gave 17,576,000 as the answer; that is not
a probability.
8. Under the same circumstances as in the previous problem what is the chance that I
get 3 different letters along with exactly 2 different numbers? (RAL 007 has 3 different
letters and 2 different numbers; so does TML 131 for instance.) [2 marks]
Now you have to count how many plates match the conditions given. The number of
possible 3 letter sequences is
26 × 25 × 24
and the number of ways to choose 3 numbers, 2 of which are equal, is
3
× 9 = 270.
10 ×
2
Here 10 is the number of possibilities for the number you have 2 of, 32 = 31 = 3 is the
number of ways to pick 2 spots of the 3 to put those two numbers, and 9 is the number
of choices for the other number once you have picked the one you will use twice. So
the chance I want is
25 · 24 · 27
6 · 27
26 · 25 · 24 · 10 · 3 · 9
=
=
= 0.239645 ≈ 0.240.
26 · 26 · 26 · 10 · 10 · 10
26 · 26 · 10 · 10
26 · 26
One mark is the answer. The other is explanation. It’s ok just to go straight to the
decimal number from the first version of the fraction; you don’t have to cancel things.
There are other ways to do the counting of the three numbers. For instance you could
say: there are 10 choices for the first number. The second number is either the same
3
as the first or different. If it is the same then there is only 1 choice for that number
and then 9 choices for the last number. If the second number is different from the first
then there are 9 choices for that number and then the last number can be either one
of the 2 digits already chosen. This adds up to 10 × 1 × 9 + 10 × 9 = 10 × 9 × 3 = 270
as before.
Grade Sheet Midterm 1
Name:
Student Number:
1
2
2
1
3
1
4
1
5
1
6
2
7
2
8
2
Total
12
4