University:
York University
Course:
AP/ADMS 2320 | Business Statistics
Academic year:

2022
Views:

315

Pages:

4
Author:

Niklas

pre-lecture notes lecture 2 Binomial Distribution binomial experiment = n repeated trials, with each trial having 2 outcomes (success or failure) P(S) = p and P(F) = 1-p trials are independent binomial RV = number of successful outcomes b(x;n,p) = (nCx)p^x(1-p)^(n-x) n = number of trials x = number of successful outcomes p = probability of success (nC) = number of combinations n choose x (nCx) = n! / (x!(n-x)!) e.g. if n=4 and x=2: (4C2) = 4! / (2!(4-2)!) -> 4x3x2x1 / (2x1x2x1) -> 24 / 4 = 6 *anything choose 0 = 1 *x choose 1 = x coin example = coin is flipped 5 times, calculate probability distribution where head is success. b(x; 5, 0.5) = (5Cx)p^x(1-p)^(n-x) P(x=0) = (5C0)0.5^0(0.5)^5 = 1x1x0.03125 = 0.03125 P(x=1) = (5C1)0.5^1(0.5)^4 = 5x0.5x0.0625 = 0.15625 P(x=2) = (5C2)0.5^2(0.5)^3 = 10x0.25x0.125 = 0.3125 P(x=3) = (5C3)0.5^3(0.5)^2 = 10x0.125x0.25 = 0.3125 P(x=4) = (5C4)0.5^4(0.5)^1 = 5x0.0625x0.5 = 0.15625 P(x=5) = (5C5)0.5^5(0.5)^0 = 1x0.03125x1 = 0.03125 mean = μx = np eg. coin example: μx = 5x0.5 = 2.5 variance σ² = np(1-p) eg coin example: σ²x = 2.5x0.5 = 1.25 binomial proportion = number of successes / number of trials: p̂ = x/n μp̂ = p σ²p̂ = p(1-p)/n Textual Content: example: p=0.3 n=3 x=2 [P(x=2)=\binom{3}{2}0.3^{2}(0.7)^{1}=3\times0.09\times0.7=0.189] at least 1 customer = p(x=1) + p(x=2)+p(x=3) [P(x=1)=\binom{3}{1}0.3^{1}(0.7)^{2}=3\times0.3\times0.49=0.441] [P(x=3)=\binom{3}{3}0.3^{3}(0.7)^{0}=1\times0.027\times1=0.027] \[P(x\ge1)=0.657\] [P(x=0)=\binom{3}{0}0.3^{0}(0.7)^{3}=1\times1\times0.343] *another way [=1-P(x=0)] Normal Distribution continuous distribution bell shaped, symmetrical N(x; μ, σ) = (1 / √(2πσ²)) * exp(-(x-μ)² / (2σ²)) when μ=0 and σ=1 = standard normal distribution SND RV: Z values = z probabilities can't be calculated for single value, must be an interval -need to calculate 2 values [Z=\frac{X-\mu}{\sigma}] [Z(sample)=\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}] example: time taken by bus = X, mean = 32 mins, var = 100 mins bus is late if it takes over 37 mins. What is prob bus is late? [Z=\frac{37-32}{\sqrt{100}}=\frac{5}{10}=0.5] [P(X>37)=P(Z>0.5)=[1-P(Z\le0.5)]\approx1-0.6915=0.3085] P(late) = bus is late 31% of times B) how long does slowest 20% of journeys last? -> all times greater than x₀ -> P(X>x₀) = 0.2 so P(X≤x₀) = 0.8 -> on z table, 0.8 = 0.84 -> so, (x₀-32)/10 = 0.84 -> 10(0.84)+32=40.4 -> 20% of slowest journeys are at least 40.4 mins