BioChemistry > STUDY GUIDE > BIOTECH CH009IU Week 4 – Lecture 3 (All)

BIOTECH CH009IU Week 4 – Lecture 3

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Continuous Distribution If we know the mean and the standard deviation, we know the distribution. - normal dis = determining the mean and sd. - n: number or trial. - p: probability of getting succ ... ess. Finding Values of a Normal Random Variable, Given a Probability The normal distribution with  = 5.5 and  = 1.6583 is a closer approximation to the binomial with n = 11 and p = 0.50 Solution: - n = 11 - p = 0.5 - mean = n*p = 5.5 =  - s.d = sqrt(n*p*q)  q=1-p  s.d = sprt(11*0.5*0.5) = 1.658312395 = . * Excel calculate Binormal: =BINOM.DIST(4,11,0.5,1)  P(X<=4) = 0.274414. (right). BINOM.DIST(P(x), n, p,1). * Excel: convert binorm to normal dis: =NORM.DIST(4,5.5,1.658) = 0.182856 (left  caclculate of thr region area). NORM.DIST(P(x), mean,s.d).  P(X<=4). - They are not same so we need to do a small adjustment. For the binomial, calculate P(X<=4) + P(X>5) * P(X>5) = 1 – NORM.DIST (5,5.5,1.658,1) = 0.618488.  P(X<=4) + P(X>5) = 0.182856 + 0.618488 = 0.801344. - With normal dis we fine things between 4 & 5 OR P(4.5): =NORM.DIST(4.5,5.5,0.1658,1) = 0.273241. Notice: For continuous dis P(X<=A) = P(X<A) = NORM.DIST(P(x), mean,s.d). P(X>A) = 1- NORM.DIST(P(x), mean,s.d). Slide 24 Mean = 124, sd = 12 Yellow part mean: X >139.36 is 0.01 or 1%  X < = 139.6 is 0.99 or 99%. Example: Mean = 0.25, sd = 0.06, calculate P(X < 0.13) and P(X > 0.31) P(X < 0.13) = NORM.DIST(0.13, 0.25, 0.06, 1). P(X > 0.31) = 1 – NORM.DIST (0.31, 0.23, 0.06, 1). Slide 23 The bell this is called the standard normal distribution that is how far you are from the mean. sd at 0.13 is (0.25 – 0.13)/0.06 = 0.12/0.06 = 2  At 0.13 is 2 sd. sd at 0.31 is (0.31 – 0.25)/0.06 = 1  At 0.31 is 1 sd. Apply the bell, at 0 = mean = 0.25.  X = 0.13 mean X at -2 and X = 0.31 mean X at 1. * Formular calculate: NORM.S.DIST (at SD,1) = NORM.S.DIST (-2,1) [P(X < 0.13)]. And NORM.S.DIST (1,1) [P(X > 0.3 [Show More]

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