2. Suppose you are to choose among two suppliers of a certain chemical solution used in paint production. The production process yields the best results if the solution has 12% concentration. Both suppliers are capable of producing the desired concentration, however there is naturally occurring random fluc- tuations in the end product. Your analysis of sales samples provided concludes that the solution concentration from Supplier 1 follows a distribution with the density function fi(r) = for 0 < a < 24, whereas the solution concentration from Supplier 2 follows distribution with the probability density function f2(x) = for 0

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2. Suppose you are to choose among two suppliers of a certain chemical solution used in paint production.
The production process yields the best results if the solution has 12% concentration. Both suppliers
are capable of producing the desired concentration, however there is naturally occurring random fluc-
tuations in the end product.
Your analysis of sales samples provided concludes that the solution concentration from Supplier 1
follows a distribution with the density function fi(x) = for 0 < a < 24, whereas the solution
2x
324
concentration from Supplier 2 follows distribution with the probability density function f2(x) =
for 0 <a < 18.
a) Suppose that a solution with concentration lower than 2% is unusable. If your goal is to minimize
the risk of receiving unusable product, which supplier should you choose?
b) If your goal is to make sure that the solution purchased has the smallest deviation from to the
target value of 12% concentration in the long run, which supplier should you choose? Explain your
answer.
Transcribed Image Text:2. Suppose you are to choose among two suppliers of a certain chemical solution used in paint production. The production process yields the best results if the solution has 12% concentration. Both suppliers are capable of producing the desired concentration, however there is naturally occurring random fluc- tuations in the end product. Your analysis of sales samples provided concludes that the solution concentration from Supplier 1 follows a distribution with the density function fi(x) = for 0 < a < 24, whereas the solution 2x 324 concentration from Supplier 2 follows distribution with the probability density function f2(x) = for 0 <a < 18. a) Suppose that a solution with concentration lower than 2% is unusable. If your goal is to minimize the risk of receiving unusable product, which supplier should you choose? b) If your goal is to make sure that the solution purchased has the smallest deviation from to the target value of 12% concentration in the long run, which supplier should you choose? Explain your answer.
Axloms of Probablity
Also Note
1. P(8)-1
2. For any event E, 0S P(E)s1
For any two events A and B,
P(A) - P(AN B) + P(ANB)
3. For any two mutually exclusive events,
and
P(EUF) - P(E) + P(F)
P(AN B) - P(A|B)P(B).
Addition Rule
Events A and B are Independent if:
P(EUF) = P(E) + P(F) - P(En F)
P(A|B) = P(A)
Conditional Probablity
or
P(B|A) -
P(ANB) - P(A)P(B).
PLAN)
Bayes' Theorem:
Total Probablity Rule
P(A|B)P(B)
P(B|A) = PLALBPB) + P(AB)P(B")
P(A) - P(A|B)P(B) + P(A|B')P(B')
Similarly,
Similarly,
P(A) -P(A|E,)P(E)) + P(A|E)P(E)+
...+ P(A|E)P(E)
P(B|E)P(E)
P(E|B) - PIBIE PE) + P(BE PE)+...+ P(B\E)P(E.)
Probability Mass and Density Functions
If X is a discrete r.v:
Cumulative Distribution Function
• F(z) = P(X sz)
P(X = 2) = f(z)
• lim,- F() -0
Es(2) =1 (total probability)
• lim,e F(z) = 1
If X is a continuous r.v.:
P(X = z) = 0
• F(z) = " /(v)dy if X is a contimuous r.v.
S(2)dz =1 (total probability)
• F(z) = E,sz f(z) if X is a discrete r.v.
• P(a < X Sb) - F(b) – F(a)
Expected Value and Variance
Expected Value of a Function of a RV
• E[X) = E, z/(z) if X is a discrete r.v.
• E[h(X)] =E. h(x)f(x) if X is a discrete r.v.
• Eh(X)) = h(z)/(z)dz if X is a continu-
• E[X] = z/(z)dr if X is a continuous r.v.
ous r.v.
• Var(X) = E[Xx] – E[X]?
• E(aX + 6) = aEX] + 6
• Var(aX + b) = a?Var(X)
%3D
• Var(X) = E[(X - E[X])?]
Derivatives and Integrals of Common Functions
• = aea
de
• Sea" dz =
• Sre*dr = e"I- fe*dz = ze" - e (using integration by parts)
dinz
• S !dz = In(z)
Common Discrete Distributions
• X - Bernoulli(p),
if z = 1;
f(z) =
|1-p ifz 0' EX] = p, Var(X) = p(1 – p).
• X- Geometric(p),
f(2) = (1– p)--'p, z E {1,2,..}, E[X] = }, Var(X) = .
Geometric Series: Eg = , for 0 < q < 1
• X - Binomial(n, p),
f(z) = (E) (1– p)"-p*, I € {0, 1,.., n},
E[X] = np, Var(X) = mp(1 – p).
%3D
• X- Negative Binomial(r, p),
f(z) = ()(1 – p)*-"p", E[X] = ;, 1 € {r,r+1,..}, Var(X) = p),
%3D
• X - Hypergeometric(n, M, N),
f(z) =
,
E[X] = n, Var(X) = N=n(1-).
%3D
• X ~ Poisson(At),
f(z) = A0", z e {0, 1, .}, E[X] = At, Var(X) = At.
Common Continuous Distributions
• X - Exponential(A),
f(z) = de-A, z E [0, 00) E[X] = }, Var(X)= .
• X- Erlang(r, A),
f(z) = A' , zE (0, 00), E[X] = 5, Var(X) = .
Suppose that Duke Energy mu
Transcribed Image Text:Axloms of Probablity Also Note 1. P(8)-1 2. For any event E, 0S P(E)s1 For any two events A and B, P(A) - P(AN B) + P(ANB) 3. For any two mutually exclusive events, and P(EUF) - P(E) + P(F) P(AN B) - P(A|B)P(B). Addition Rule Events A and B are Independent if: P(EUF) = P(E) + P(F) - P(En F) P(A|B) = P(A) Conditional Probablity or P(B|A) - P(ANB) - P(A)P(B). PLAN) Bayes' Theorem: Total Probablity Rule P(A|B)P(B) P(B|A) = PLALBPB) + P(AB)P(B") P(A) - P(A|B)P(B) + P(A|B')P(B') Similarly, Similarly, P(A) -P(A|E,)P(E)) + P(A|E)P(E)+ ...+ P(A|E)P(E) P(B|E)P(E) P(E|B) - PIBIE PE) + P(BE PE)+...+ P(B\E)P(E.) Probability Mass and Density Functions If X is a discrete r.v: Cumulative Distribution Function • F(z) = P(X sz) P(X = 2) = f(z) • lim,- F() -0 Es(2) =1 (total probability) • lim,e F(z) = 1 If X is a continuous r.v.: P(X = z) = 0 • F(z) = " /(v)dy if X is a contimuous r.v. S(2)dz =1 (total probability) • F(z) = E,sz f(z) if X is a discrete r.v. • P(a < X Sb) - F(b) – F(a) Expected Value and Variance Expected Value of a Function of a RV • E[X) = E, z/(z) if X is a discrete r.v. • E[h(X)] =E. h(x)f(x) if X is a discrete r.v. • Eh(X)) = h(z)/(z)dz if X is a continu- • E[X] = z/(z)dr if X is a continuous r.v. ous r.v. • Var(X) = E[Xx] – E[X]? • E(aX + 6) = aEX] + 6 • Var(aX + b) = a?Var(X) %3D • Var(X) = E[(X - E[X])?] Derivatives and Integrals of Common Functions • = aea de • Sea" dz = • Sre*dr = e"I- fe*dz = ze" - e (using integration by parts) dinz • S !dz = In(z) Common Discrete Distributions • X - Bernoulli(p), if z = 1; f(z) = |1-p ifz 0' EX] = p, Var(X) = p(1 – p). • X- Geometric(p), f(2) = (1– p)--'p, z E {1,2,..}, E[X] = }, Var(X) = . Geometric Series: Eg = , for 0 < q < 1 • X - Binomial(n, p), f(z) = (E) (1– p)"-p*, I € {0, 1,.., n}, E[X] = np, Var(X) = mp(1 – p). %3D • X- Negative Binomial(r, p), f(z) = ()(1 – p)*-"p", E[X] = ;, 1 € {r,r+1,..}, Var(X) = p), %3D • X - Hypergeometric(n, M, N), f(z) = , E[X] = n, Var(X) = N=n(1-). %3D • X ~ Poisson(At), f(z) = A0", z e {0, 1, .}, E[X] = At, Var(X) = At. Common Continuous Distributions • X - Exponential(A), f(z) = de-A, z E [0, 00) E[X] = }, Var(X)= . • X- Erlang(r, A), f(z) = A' , zE (0, 00), E[X] = 5, Var(X) = . Suppose that Duke Energy mu
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