Unformatted text preview: = 245, p = 0.25) ≈ N (µ = 61.25, σ = 6.78).
0.06
Bin(245,0.25)
N(61.5,6.78) 0.05 P (X ≥ 70) = P (X = 70 or X = 71 or X = 72 or · · · or X = 245) 0.04 = P (X = 70) + P (X = 71) + P (X = 72) + · · · + P (X = 245) 0.03
0.02 This seems like an awful lot of work... 0.01
0.00
20 Sta102/BME102 (Colin Rundel) Lec 6 February 5, 2013 15 / 23 Sta102/BME102 (Colin Rundel) 40 60 Lec 6 k 80 100 February 5, 2013 16 / 23 Normal Approximation to the Binomial Basics Normal Approximation to the Binomial Improvements Facebook cont. Improving the approximation What is the probability that the average Facebook user with 245 friends
has 70 or more friends who would be considered power users? Take for example a Binomial distribution where n = 20 and p = 0.5, we should be able
√
to approximate the distribution of X using N (10, 5). 2 3 4 5 6 7 8 9 10 12 14 16 18 It is clear that our approximation is missing about 1/2 of P (X = 7) and P (X = 13), as
n → ∞ this error is very small. In this case P (X = 7) = P (X = 13) = 0.073 so our
approximation is oﬀ by ≈ 7%.
Sta102/BME102 (Colin Rundel) Lec 6 Normal Approximation to the Binomial February 5, 2013 17 / 23 Improvements Sta102/BME102 (Colin Rundel) Lec 6 Normal Approximation to the Binomial February 5, 2013 18 / 23 Improvements Improving the approximation, cont. Improving the approximation, cont. Binomial probability: This correction also lets us do, moderately useless, things like calculate the
probability for a particular value of k . Such as, what is the chance of 50
Heads in 100 tosses of slightly unfair coin (p = 0.55)? 13 P (7 ≤ X ≤ 13) =
k =7 20
0.5k (1 − 0.5)20−k
k Binomial probability: Naive approximation: P (X = 50) = P (7 ≤ X ≤ 13) ≈ P Z≤ 13 − 10
√
5 −P Z≤ 7 − 10
√
5 Naive approximation:
P (X = 50) ≈ P Z ≤ Continuity corrected approximation:
P (7 ≤ X ≤ 13) ≈ P Z≤ 13 + 1/2 − 10
√
5 −P Z≤ 7 − 1/2 − 10
√
5 Lec 6 February 5, 2013 50 − 55
4.97 −P Z ≤ 50 − 55
4.97 =0 Continuity corrected approximation:
P (X = 50) ≈ P Z ≤ Sta102/BME102 (Colin Rundel) 100
0.5550 (1 − 0.55)50 = 0.04815
50 19 / 23 Sta102/BME102 (Colin Rundel) 50 + 1/2 − 55
√
4.97 −P Z ≤ Lec 6 50 − 1/2 − 55
√
4.97 = 0.04839 February 5, 2013 20 / 23 Normal Approximation to the Binomial Improvements Normal Approximation to the Binomial Improvements Example  Rolling lots of dice Example  Airline booking Roll a fair die 500 times, what’s the probability of rolling at least 100 ones? An airline knows that over the long run, 90% of passengers who reserve
seats show up for ﬂight. On a particular ﬂight with 300 seats, the airline
accepts 324 reservations.
If passengers show up independently what is the probability the ﬂight will
be overbooked?
Suppose that people travel in groups, does this increase or decrease the
chance of overbooking? Sta102/BME102 (Colin Rundel) Lec 6 Normal Approximation to the Binomial February 5, 2013 21 / 23 Improvements Example  Voter support
Suppose 55% of a large population of voters support actually favor a
particular candidate. How large a random sample must be take for there
to be a 99% chance that the majority of voters in the sample will favor
that candidate? Sta102/BME102 (Colin Rundel) Lec 6 February 5, 2013 23 / 23 Sta102/BME102 (Colin Rundel) Lec 6 February 5, 2013 22 / 23...
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 Spring '08
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 Statistics, Binomial, Biostatistics, Normal Distribution, Probability, Binomial distribution, Colin Rundel

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