Jaclyn Ebert 6221545
Probability time segments. Per – purr – cat – cat eats fish – poisson. Describes
the # of events in a given interval of time or space.
Marginal (straight): P(x) -λ x
P(x∩y) = P(x AND y) = P(x)*P(y) P(x) = e (λ )/x!
P(0 OR 1 OR 2 OR 3) = e (1/1 + λ /1! + λ /2! + λ /3!)
P(xuy) = P(x OR y) = P(x) + P(y) – P(x AND y) P(>3) = 1 – P(0 OR 1 OR 2 OR 3)
Mutually exclusive (disjoint): if the occurrence of one event Mean of poisson = μ = λ do rate*units… ex: rate 0.6, takes 1.7
precludes the occurrence of another event. hours, so 0.6*1.7=1.02 and that is your lamda
Then P(x OR y) = P(x) + P(y)
SD of poisson = σ = √λ
Conditional: P(x|y) = P(x given that y has occurred)
P(xANDy) Exponential: Interval of time or space between events. Probability
= /P(y) where P(y) > 0 density function.
Independent: when P(y|x) = P(y) -λx
f(x) = λe where x ≥ 0
or P(x AND y) = P(x)*P(y)
vs. Disjoint: when P(x AND y) = 0 or when A and B overlap and Standard Normal Distrubution
when both events can occur together.
I know P(x|y), and I want to know P(y|x).
I have an initial estimate of P(y), I do a test and find that x is true, so I
update my initial estimate to get P(y|x). Use tree diagrams, then
multiple along the branches to get PAND y). Then to find P(y|x), use
the conditional formula, and the denominator will be all the P(y)s μ x
z = x-μσ where: x = any point on the horizontal axis, σ = SD of
P(D|+) = P(+|D)*P(D) / [ P(+|D)*P(D) + P(+|W)*P(W) ]
Where: + = positive test result, D = patient has diabetes, W the normal distribution, μ = population mean. Then use table.
= patient is well. Backwards: z = find P in middle of table somewhere, then:
Random Variables x = μ + zσ
Random variables (RVs) ‘X’ and ‘Y’ represent annual incomes of well
E(x) = EV = mean = μ = ∑x*P(x) established unmarried young professional men and women. These
two RVs are normally distributed and are statistically independent.
Var(x) = σ² = ∑(x – EV)²[P(x)] = ∑[x – x*P(x)]² [P(x)]
SD(x) = σ = √Var(x) = √∑[x – x*P(x)]² [P(x)] The mean values of ‘X’ and ‘Y’ are $100K and $80K with standard
deviations of $20K and $15K respectively.
Properties: Find P(The woman has a higher income than that of the man).
E(x+y) = E(x) + E(y) P(W>M) = P(W-M>0)
P(W-M>0) μ SD Var
Var(x+y) = Var(x) + Var(y) W 80 15 ² => 225 ↓
SD(x+y) = √[Var(x) + Var(y)]
M 100 20 ² => 400 +
Discrete Probability Distrubutions W-M = -20 25 <= √ 625
Uniform: equal probabilities for all possible outcomes. Ex: roll z = (x-μ)/σ = (0+20)/25 = 0.8 P = 0.7881 from table
P(W>M) = 1 – 0.7881 = 0.21
dice. Use random variable table.
Combinations: Select x out of n with result good or bad. Correlation Coefficient:
C = / x!*(n – x!) r = ∑(x-)(y-) s x / ∑(x-)² = ∑x /in
(n-1)(s )(s ) √ n-1
Binomial: x out of n, two possible outcomes with n identical and x y
independent trials. P of success remains constant from trial to trial. Correlation between Random Variables: Corr(x,y) ∑(x-)(y-)p
Sampling with replacement. Known probability of being special.
P(x) = /n! x!*(n-x)!)(q )-x (sx)(s y
where: n = sample size, x = number of successes, n-x = number Joint Probabilities Meter: $1 Meter: $2
of failures, p = probability of success, q = (1-p) = probability of Fine: $0 0.5 0.4
failure. Fine: $50 0.01 0.09
Mean of binomial = μ =xnp x P(x) Mean Variance
SD of binomial = σ = √npq = √np(1-p) 1 0.5+0.01=0.51 1(0.51)=0.51 (1-1.49)²(0.51)=0.12245
Normal Approximation: If np>10 and n(1-p)>10, then if P(≥50), 2 0.4+0.09=0.49 2(0.49)=0.98 (2-1.49)²(0.49)=0.1274
subtract 0.5 from x in the z calculation, and if P(≤50), add 0.5.
ex: μ = 10 for a normal distribution, σ = 2.83 s=√Var=0.5
P(10) = P(9.5≤x≤10.5) = P[(9.5-10)/2.83≤z≤(10.5-10)/2.83] = P(-
y P(y) Mean Variance
0.177≤z≤0.177) = 0.140( - μ) N-n yP(y) (y-)²P(y)
Finite population: z = /[σ√( / N-1 0 0.5+0.4=0.9 0(0.9)=0 (0-5)²(0.9)=22.5
50 0.01+0.09=0.1 50(0.1)=5 (50-5)²(0.1)=202.5
Poisson: Known arrival rate. Average # of events per segment is λ. =5 Variance=225
# of events is random, occurrence of one does not influence the
chance an other. Probability that an event occurs in a given segment s=√Var=15
is the same for all segments. Ex: # customers returning products per Corr(x,y) = [(1-1.49)(0-5)0.5 + (2-1.49)(0-5)0.4 +
month, # phone calls per min, # cracks per km of highway. Usually (1-1.49)(50-5)0.01 + (2-1.49)(50-5)0.09] / (0.5*15)Normal
The answer will
No be approximate
c (4 marks You want to examine 20 market-size cod which have achieved a weight of over 4.25 kg
without your new feeding program. You take a random sample of 250
market-sized cod. What is the
probability that this sample will contain enough cod for your purposes?
market-size, with a standard deviation of 1 month. The average weight of market size cod is 3.75 kg
with a standard deviation of 0.38 kg, and we wil assume it is Normally distributed. [denived from:
Normal opulation Yes Normal Dist Yes Large Dist Sample? The answer will No be approximate c (4 marks You want to examine 20 market-size cod which have achieved a weight of over