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STAT 10-12-16

Binomials (Test 1)

x~b(40, .6), P(x < 30) = P(x <= 29) = BINOMDIST(29,40,0.6,1)
x~b(40, .6), P(x > 28) = 1 - P(x <= 28) =1-BINOMDIST(28,40,0.6,1)
x~b(40, .6), P(20 < x < 30) = P(x <= 29) - P(x <= 20) =BINOMDIST(29,40,0.6,1)-BINOMDIST(20,40,0.6,1)
x~b(40, .6), P(20 > x > 24) = 1 - [P(x <= 24) - P(x <= 19)] =1-BINOMDIST(24,40,0.6,1)+BINOMDIST(19,40,0.6,1)
μ mean
σ2 variance
σ standard deviation
Suppose x~ b (n, p). Since x is discrete, μ = ?, σ2 = ? np; np(1-p)
x~b(40, .6), find mean and variance μ = np = 40(.6) = 24, σ2 = np(1-p) = 40(.6)(1-.6) = 9.6
Normal approximation to the binomial in some situations, it is possible to approximate binomial probabilities using the normal distribution
Why approximate probabilities when we can find them exactly? binomial probabilities are hard to calculate without a computer, and the normal distribution can be used to calculate probabilities for functions or binomial random variables
x~b(n,p), then x/n = successes/trials
If x is approximately normal, then p^ is too
Normal approximation to the binomial in Minitab Graph → Probability distribution plot → Two distributions
Standard deviation square root (variance)
Rule of thumb is we want both np and n(1-p) to be away from zero
Usually, “away” means at least 5 (some textbooks at least 10)
For the normal distribution, what do we need to calculate probabilities? mean and standard deviation
From a binomial probability question, find the mean and standard deviation and pretend it’s a normal probability question
x~b(40, .6), P(x < 30) = 0.964 (exact). np=24, n(1-p)=16 = good, P(x<30) is approximately =NORMDIST(30,24,3.098,1)
A better approximation is to use the continuity correction factor (splitting the difference)
Created by: Spencer Gowey Spencer Gowey