Chapter 1 & 2
Give three reasons why we use sample instead of using the whole population?
1) sampling is faster and less expensive
2) samples are much smaller and easier to analyze
3)sometimes a full survey of the entire population is impossible
What is the difference between the following approaches to probability: Theoretical, Relative Frequency
Theoretical Approach- use math and first principles to derive a formula for the probability of an event.
Relative Frequency- examine past results to come up with a proportion that should approximate the probability.
Suppose X~N( -2, 0.5), what is the value of P(X= -2)?
0.5000
What are the four principles of experimental design? And be able to provide a short definition of each.
Controlling- minimizing differences between individuals by providing clear instructions.
Randomization- prevents researchers from influencing results.
Replication- the more data you collect, the more confident you will be in your results.
Blocking- divide participants into groups with similar characteristics.
Suppose P(A)= 0.4, P(B)= 0.8 and P(A and B)= 0.3 determine whether A and B are independent.
P(A and B)=0.3= P(A)P(B)
(PA)P(B)= (0.4)(0.8)= 0.32
0.32≠0.3
A and B are dependent
Suppose X~N( -2, 0.5), what is the value of Var(X)?
0.25
Find the variance of: 2 0 1 2 25
formula: s2= Σ (x1-x̄ )2
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n-1
Solution: x̄ = 0+1+2+2+25 = 30/5 = 6
s2=(0-6)2+(1-6)2+(2-6)2+(2-6)2+(25-6)2
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5-1
= (-6)2+(-5)2+(-4)2+(-4)2+(19)2
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4
=454/4= 113.5
What is the probability of drawing a 6 or a heart from a deck of cards?
There are 4 six and 13 hearts in a deck. However, there is 1 six of hearts that overlaps.
P(six or heart)= P(six) + P(heart) - P(six and heart)
P(six or heart)= 4/52 +13/52 - 1/52
P(six or heart)= 16/52 --> 4/13 =0.3077
Suppose X~N(-8, 4), determine the P(X<2)?
Solution:
Z = X − µ = 2 − (−8) = 10 = 2.5
-------- ---------- ----
σ 4 4
B) Find the quartiles of: 1 , 4 , 2 , 8 , 5 , 0 , 4 , 6 , 3
The three quartiles are:
Q1= the median of the lower half of the data set.
Q2= the median of the entire data set collected.
Q3= median of the upper half of the data set.
Solution to part B: (note: first put the numbers in order)
0 , 1 , 2 , 3 , 4 , 4 , 5 , 6 , 8
Q1= 0 , 1 , 2 , 3
Q2= 4
Q3= 4 , 5 , 6 , 8
What is the probability of rolling 6 dice and getting at least one 3?
Let B be "at least one 3". Bc is "zero 3s"
P(Bc)= (5/6)(5/6)(5/6)(5/6)(5/6)(5/6)= 15625/46656 or 0.3348979767
P(B)= 1-15625/46656 OR 1-0.3348979767
P(B)= 0.6651020233
Q: A cat has a litter of 7 kittens, each kitten has an independent 1\3 chance of having black fur. Let X be the number of kittens with black fur. Determine whether or not the X is a binomial random variable.
X is a binomial random variable because:
-the question says the trials are independent
-there are 7 trials (kittens) n=7
- succes is black fur, failure is anything else
- probabilty of succes is always p=1\3
A) What is the five number summary and what purpose does it serve in statistics?
B) Write out a five number summary for the following numbers:
9 , 7 , 2 , 0 , 6 , 11 , 9 , 8 , 13 , 4 , 11
A) The five number summary is a statistic of five numbers including: the minimum, Q1, Q2, Q3, and the maximum. The purpose it serves in statistics is that it gives information about the center, variation and skew of a distribution.
Solution to part B)
(*note to allows put the numbers in order first)
0 , 2 , 4 , 6 , 7 , 8 , 9 , 9 , 11 , 11 , 13
The five number summary is:
0 , 4 , 8 , 11 , 13
Q: A weighted 8-sided die has a 0.1 chance of rolling a 1, a 0.2 chance of rolling a 2, a 0.3 chance of rolling a 3 and a 0.0.8 chance of rolling the other numbers (4, 5, 6, 7, 8). What is the probability of rolling an even number on this die?
P(even)= P(2 or 4 or 6 or 8)
P(even)= P(2)+P(4)+P(6)+P(8)
P(even)= 0.2 + 0.08 + 0.08 + 0.08
P(even)= 0.44
Q: In Canada 46% people have 0-type blood. Let X be the number of people with 0-type blood in a random sample of 10 Canadians. Confirm that X is a binomial random variable and calculate E(X), SD(X), P(X ≤ 1) and P(X ≥2).
X is a binomial random variable because:
-trials are independent (random sampling from a large population)
-there are 10 trials
-0-type blood is a succes, anything else is a faliure
-probability of succes is p=0.46 (46%)
E(X)=p*n=(10)(0.46)=4.6
SD(X)=√var(x)=√(10)(0.46)(0.54)=1.5761
P(X ≤ 1) =P(X=0 or X=1)=P(X=0)+P(X=1)
=(10)(0.46)0(0.54)10+(10)(0.46)1(0.54)9=0.0021+0.018
=0.0201
P(X ≥2)= 1-P(X ≤ 1)= 1-0.0201= 0.9799