This is an ungraded and anonymous survey for me to evaluate the distribution of your math and statistics backgrounds. Please write your answers the the following questions on a piece of paper, or download and print a PDF copy (link above) to write on. You do not need to write your name on it. Please complete all problems to the best of your ability. Your responses will help me craft the course to see which material we need to focus on at greater length, especially review material.
8,12,9,10,11,5,15
First, arrange these in numerical order. 5,8,9,10,11,12,15
ˉx=1nn∑i=1xiˉx=5+8+9+10+11+12+156=707=10
Sample standard deviation is the average deviation of an observation from the mean, calculated as the square root of variance (which is the average squared deviation from the mean), (with a slight correction for samples: n−1 rather than n). First we find the variance:
s2=1n−1n∑i=1(xi−ˉx)2
You can calculate this all at once: s2=(5−10)2+(8−10)2+(9−10)2+(10−10)2+(11−10)2+(12−10)2+(15−10)27−1=606=10
But it is often easier to first find the deviations from the mean for each data point (2nd column), square them (3rd column), add them all up (bottom row) and then divide by n−1:
xi | (xi−ˉx) | (xi−ˉx)2 |
---|---|---|
5 | −5 | 25 |
8 | −2 | 4 |
9 | −1 | 1 |
10 | 0 | 0 |
11 | 1 | 1 |
12 | 2 | 4 |
15 | 5 | 25 |
∑ni=1(xi−ˉx)2 | 60 |
s2=1n−1n∑i=1(xi−ˉx)2=17−1[60]=606=10
Standard deviation is the square root of variance: s=√s2=√10≈3.16
For a fair, six-sided die:
All outcomes {1,2,3,4,5,6} are equally likely, so the chance of any one outcome is the same, 16.
There are three possible disjoint outcomes, rolling a 2, 4, or 6, so we add up the individual probabilities: P(x=even)=P(x=2)+P(x=4)+P(x=6)=(16)+(16)+(16)=36=0.5
We know the probability of an even roll is 0.5, and rolling a 3 is disjoint from rolling an even number (a roll cannot be simultaneously 3 and even). So this is the union ∪ of two events. P(x=even OR x=3)=P(x=even)+P(x=3)=(36)+(16)=46
This is the intersection ∩ of two independent possibilities: that each die will role a six, each with the probability of 16, and that the outcome of the first die has no bearing on the outcome of the second die. Thus, we can simply multiply the two events’ probabilities:
P(x1=6,x2=6)=P(x1=6)×P(x2=6)=(16)×(16)=136
We must standardize these variables and compare their Z-scores. We calculate an observation’s Z-score relative to its distribution by taking the observation, subtracting the mean of that variable, and dividing by its standard deviation:
Z=xi−ˉXsX
For hedge fund A:
Z=4.0−2.50.5=3
A 4.0% return for Hedge Fund A is 3 standard deviations above its mean return.
For hedge fund B:
Z=−1.0−3.02.0=−2
A −1.0% return for Hedge Fund B is 2 standard deviations below its mean return.
The event in hedge fund A is more extreme.
x | p(x) |
---|---|
10 | 0.1 |
20 | 0.2 |
30 | 0.3 |
40 | 0.4 |
First we must calculate the mean, μ or E(X):
E(X)=n∑i=1xipi=0.1(10)+0.2(20)+0.3(30)+0.4(40)=1+4+9+16=30
Now we must calculate the variance, the average squared distance from the mean (30) weighted by the probability of each value. We can do it all at once:
var(X)=E[(Xi−μ)2]=∑pi(Xi−μ)2=0.1(10−30)2+0.2(20−30)2+0.3(30−30)2+0.4(40−30)2=0.1(400)+0.2(100)+0.3(0)+0.4(100)=40+20+0+40=100
Alternatively, we could use the table method to make sure we keep our numbers straight:
xi | pi | (xi−μ) | (xi−μ)2 | pi(xi−μ)2 |
---|---|---|---|---|
10 | 0.10 | −20 | 400 | 40 |
20 | 0.20 | −10 | 100 | 20 |
30 | 0.30 | 0 | 0 | 0 |
40 | 0.40 | 10 | 100 | 40 |
n∑i=1pi(xi−μ)2 | 100 |
The first column is each individual value. The second is the probability of each value. For the third row, we take the deviation of each value from the mean. The fourth column, we square it. The fifth column, we multiply the squared deviation by the probability of the value. Then, we add up the fifth column to get the variance, 100.
Standard deviation is just the square root of the variance: sd(X)=√var(X)=√100=10
Y∼N(50,12)
The z-score formula subtracts the mean from a value and divides by the standard deviation to convert any normal distribution to the standard normal distribution (Y∼N(0,1): z=y−μσz=74−5012z=2
The z-score or the standard score measures how many standard deviations a value is away from the distribution’s mean. Thus, this value, 74, is 2 standard deviations away from the distribution’s mean of 50.
This requires remembering an important property of standard normal distributions, the 68-95-99.7 rule: 68% of a normal distribution falls between 1 standard deviation from the mean, 95% between 2 standard deviations, and 99.7 between 3 standard deviations. Our z-score told us that we are 2 standard deviations from the mean, so there is a 95% chance that Y will fall within 2 standard deviations from the mean (50±12, so between 26 and 74).
However, we need to be careful, because the remaining 5% can fall on either side of two deviations from the mean (). What we care about is two deviations above the mean, that is, above 74. So divide this 5% in half, to account for the 2.5% chance that Y takes on a value greater than 74.
Hint: Standardize the two hedge funds.↩︎