Download Answers for Practice Test 2

Practice Test
For the purposes of this exam, recall that 1 Watt = 1 Joule/sec, and that a kilowatt is 1,000
watts.
1. We are told that about 81,000× 1012 W of the sun’s energy reaches the ground on the
earth. How many joules is that in one day?
Answer: One Watt is one joule per second. The question is basically asking, this is how many
joules we are getting per second. Now how many joules is that per day? So you multiply the
number of Watts by number of seconds in a day. In other words, 81,000 × 1012 𝑡𝑖𝑚𝑒𝑠 60 x
60 x 24.
2. If 1 metric tonne of oil (1 toe) has 11,628 kWh of energy, find out how many joules
11,433,918 ktoe represents. (remember k = kilo = 103.)
Answer: First of all, ktoe means kilo ton oil equivalent. Kiloton means 1000 tons.
One ton is 11,628 Kwh. So one kiloton is 11,628,000 Kwh.
So 11,433,918 ktoe is 11,433,918 x 11,628,000 KWh.
To find out how many joules it is, remember that one Kilowatt-hour (KWh) is 1000 Watthours. One Watt is one joule per second. One Watt-hour is 60 X 60 or 3600 Joules because
there are that many seconds in an hour. One Kilowatt-hour is 3600 x 1000 Joules.
So 11,433,918 ktoe is 11,433,918 x 11,628,000 x 3600 x 1000 Joules.
After multiplying it is 4.786 x 1020 Joules.
3. If the figure 11,433,918 ktoe is the amount of energy consumed globally in 2005, what
was the average power consumption in terawatts (1 TW = 1012 watts)? (You must find
the number of seconds in a year.) You may use your answer from 2 to find this.
Answer: We saw that it is 4.786 x 1020 Joules from problem 2. This is for the whole year.
We need to see how many joules it is per second because one Watt is one Joule per second.
That is why we need to divide by number of seconds in a year. This is 60 X 60 x 24 x 365.
Dividing 4.786 x 1020 by this number of seconds we get that 1.5 x 1013 Watts were consumed on
average. Now one terawatt is 1012 watts, so we divide this by 1012 to see how many terawatts
this is. We get 1.5 x 10 or 15 terawatts. (Dividing powers of 10 – just subtract the exponents).
4. Centralized vs. modular method. A big, central generating plant producing 100 MW of
power takes 10 weeks to build and costs $1 million. A small, modular generating plant
producing 10 MW of power takes 1 week to build and costs $100,000.
a. What is the cost of building 100 MW of production, using the central method?
Using the modular method?
Answer: 1 million for both. In modular case you multiply 100000 by 10.
b. Two jurisdictions begin building their generators on the same day. One will
build 1 central 100 MW unit; the other will build 10 modular 10 MW units. How
much power capacity does each method have?
Answer: 100 MW for both. In modular case multiply 10 MW by 10.
c. Tell the amount of energy produced by each method during each of these time
periods (there are 168 hours in a week):
Time period
Central method
Modular method
st
1 week
0
0
2nd week
0
10 MW× 168 ℎ = 1680 MWh
rd
3 week
0
20 x 168
4th week
0
30 x 168
th
5 week
0
40 x 168
6th week
0
50 x 168
th
7 week
0
60 x 168
8th week
0
70 x 168
th
9 week
0
80 x 168
10th week
0
90 x 168
th
11 week
100 x 168
100 x 168
The reason each week the amount of energy from modular method increases by
10 Is because another powerplant with 10 mw capacity is added.
d. What is the total amount of energy produced by each method after 10 weeks?
Answer: The central method adds upto 0.
The modular method adds up to (10+20+…+90)x 168.
Now 10+20+30+…+80+90 = 10x1 + 10x2 + 10x 3 + …+ 10x8 + 10 x 9
= 10(1+2+3+…+8+9) = 10 ( (9 x 10 ) /2) = 450.
Here we used the formula 1+2+3+…+(n-1)+n = (n x (n+1) ) /2.
[In other words, to add the first n numbers, multiply the last one by itself plus one, then
divide by 2].
So total energy by modular method after 9 weeks equals 450 x 168 MWh.
e. What is the total amount of energy produced by each method after 11 weeks?
Answer: In the eleventh week the central method has 16800 MWh from the first 100 MWh
plant coming into action.
In the modular method using the same formula as in d (with 11 instead of 10) we get (10+
20+….+100) x 168 = 550 x 168.
5. Five cards are drawn from a standard 52-card deck. Find the probability that at least
one card is a face card.
Answer: You can find the probability of getting at least one face card , say P( f > 0 ) by adding
the probabilities of getting exactly one face card, exactly two face cards, and so on upto
exactly 12 face cards. (There are 12 face cards total: King, Queen Jack with four designs
each). In other words, P(f > 0) = P(f = 1) = P(f = 2) + … + P(f = 12).
But this is too long. Here is the short way:
P(f > 0 ) = 1 – P (f = 0).
In other words, P( at least one face) = 1 – P(no face card).
This uses the rule that P(event) = 1 - P(opposite event).
For example, if P(rain) = 30% = 0.3, then P(no rain ) = 70% = 0.7 = 1 – 0.3.
Now P(no face card) = Number of favorables / Total number of possibilities
= Number of ways to pick 5 non-face cards / Number of ways to pick 5 out of 52
= 40C5 / 52C5 = (40x39x38x37x36 / 5x4x3x2x1) / (52x51x50x49x48 / 5!)
= 0.253.
So P( at least one face card) = 1 – 0.253 = 0.747.
6. A couple decides to have four children. Assume that having a boy or girl is equally likely.
a. List the sample space of this experiment.
b. Find the probability that the couple has only boys.
c. Find the probability that the couple has two boys and two girls.
d. Find the probability that the couple has four children of the same sex.
e. Find the probability that the couple has at least two girls.
Answer: (a) List the number of ways they can have 4 children. Each child can be B or G. So your
answer will look like this: BBBB, GGGG, BBBG, BBGB,BGBB,GBBB, ETC.,
There are 16 total possibilities because each one is 2 and so total is 2 x 2 x 2 x 2 = 16.
This is generally the case when you have repetition possible. For example, the number of
combinations with three numbers where each one can be 0,1,2,3,…9 is 10 x 10 x 10 = 1000.
Some example combinations are 000, 001, 010, 111, 123, 134,…
(b) There is only one case with four boys: BBBB. So P(all boys) = 1/16.
(c) Count and see that answer is 6/16 = 3/8. (BBGG,BGBG,BGGB,GBGB,GBBG,GGBB).
(d) Either BBBB or GGGG so 2/16 = 1/8.
(e) Count to get 11/16.
This can also be done by the rule P(event) = 1 – P(opposite event).
Here the opposite event is getting one or no girls. P(1 G) = 4/16 because the girl could be one of
four children, with other three being boys. GBBB, BGBB, BBGB, BBBG.
P( no girls) = 1/16 ( = all boys). So totally we get 4/16 + 1/16 = 5/16. Subtracting this from 1 you
get 11/16.
7. A monkey is trained to arrange wooden blocks in a straight line. She is then given 11
blocks showing the letters A, B, B, I, I, L, O, P, R, T, Y. What is the probability that the
monkey will arrange the blocks to spell PROBABILITY?
The number of ways to spell PROBABILITY correctly is just one.
For number of ways to arrange these letters the method is similar to the one we used on the
letters of MISSISSIPPI, FREEZE etc., Here there are 11 letters with B and I appearing twice
each. So total number of arrangements is 11! / (2! X 2! ).
Divide 1 by this number to get the answer.
For the next three problems read the handout “Basic statistics with examples from
environmental studies” from Update page as well as assignment 13.
8. Make a bar graph showing the frequency distribution of the following data. These
represent the number of children in each household of a particular neighborhood: 0, 3,
4, 4, 2, 2, 1, 1, 0, 0, 3, 0, 2, 2, 1
9. Find the mean, median, and mode of these high temperatures for the Washington area
the week of October 22: 68, 67, 60, 54, 56, 68, 73
10. Find the variance and standard deviation of the temperatures in #9.