# Normal distribution Approximating binomial distribution by normal 2.10 Central Limit Theorem

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1 1.1.2 Normal distribution Approimating binomial distribution by normal 2.1 Central Limit Theorem Prof. Tesler Math 283 October 22, 214 Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

2 Normal distribution a.k.a. Bell curve and Gaussian distribution The normal distribution is a continuous distribution. Parameters: µ = mean (center) σ = standard deviation (width) ( ) PDF: f X () = 1 σ ep ( µ)2 for < <. 2π 2σ 2 Normal distribution N(2, 5): µ = 2, σ = Normal µ µ ± σ The normal distribution is symmetric about = µ, so median = mean = µ. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

3 Applications of normal distribution Applications Many natural quantities are modelled by it: e.g., a histogram of the heights or weights of everyone in a large population often follows a normal distribution. Many distributions such as binomial, Poisson,... are closely approimated by it when the parameters are large enough. Sums and averages of huge quantities of data are often modelled by it. Coverage in DNA sequencing Illumina GA II sequencing of E. coli at 6 coverage. Chitsaz et al. (211), Nature Biotechnology % of positions with coverage Empirical distribution of coverage Coverage Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

4 Cumulative distribution function It turns out that the integral for total probability does equal 1: ) 1 ( σ 2π ep ( µ)2 2σ 2 d = 1 However, it s a tricky integral; see a Calculus tetbook in the section on double integrals in polar coordinates. The cumulative distribution function is the integral F X () = P(X ) = ) 1 ( σ 2π ep (t µ)2 2σ 2 dt This integral cannot be done symbollically in terms of the usual functions (polynomials, eponentials, logs, trig functions, etc.). It can be done via numerical integration or Taylor series. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

5 Standard normal distribution Standard normal distribution N(, 1): µ =, σ = 1 CDF of standard normal distribution..2.4 Normal µ µ ± σ cdf..4.8 Normal µ µ ± σ z The standard normal distribution is the normal distribution for µ =, σ = 1. Use the variable name Z: PDF: φ(z) = f Z (z) = e z2 /2 2π for < z < CDF: Φ(z) = F Z (z) = P(Z z) = 1 z e t2 /2 dt 2π The integral can t be done in terms of ordinary functions, but it can be done using numerical methods. In the past, people used lookup tables. We ll use functions for it in Matlab and R. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36 z

6 Matlab and R commands For the standard normal: Φ(1.96).975 Φ 1 (.975) 1.96 Matlab: normcdf(1.96) norminv(.975) R: pnorm(1.96) qnorm(.975) We will see shortly how to convert between an arbitrary normal distribution (any µ, σ) and the standard normal distribution. The commands above allow additional arguments to specify µ and σ, e.g., normcdf(1.96,,1). R also can work with the right tail directly: pnorm(1.96, lower.tail = FALSE).975 qnorm(.975, lower.tail = FALSE) 1.96 Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

7 Standard normal distribution areas.4 Standard Normal Curve.3.2.1!5 a b 5 z The area between z = a and z = b is P(a Z b) = 1 2π b a e t2 /2 dt = Φ(b) Φ(a) P(1.51 Z 1.62) = Φ(1.62) Φ(1.51) = =.129 Matlab: normcdf(1.62) normcdf(1.51) R: pnorm(1.62) pnorm(1.51) Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

8 Standard normal distribution symmetries of areas.4 Area! on the right "#\$ &'()*!*+,*-.(*/( "#! z!!2 2 z "!%!!! " %! Area right of z is P(Z > z) = 1 Φ(z). By symmetry, the area left of z and the area right of z are equal: Φ( z) = 1 Φ(z) Φ( 1.51) = 1 Φ(1.51) = =.655 Area between z = ±1.51 is Φ(1.51) Φ( 1.51) = 2Φ(1.51) Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

9 Central area Area between z = ±1 is 68.27%. Area between z = ±2 is 95.45%. Area between z = ±3 is 99.73%..4.2 Area! split half on each tail!z!/2 z!/2!2 z 2 Find the center part containing 95% of the area Put 2.5% of the area at the upper tail, 2.5% at the lower tail, and 95% in the middle. The value of z putting 2.5% at the top gives Φ(z) = 1.25 =.975. Notation: z.25 = The area between z = ±1.96 is about 95%. For 99% in the middle,.5% on each side, use z Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

10 Areas on normal curve for arbitrary µ, σ P(a X b) = b a ) 1 ( σ 2π ep ( µ)2 2σ 2 d Substitute z = µ σ (or = σz + µ) into the integral to turn it into the standard normal integral: ( a µ P X µ b µ ) ( a µ = P Z b µ ) σ σ σ σ σ ( ) ( ) b µ a µ = Φ Φ σ σ The z-score of is z = µ σ. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

11 Binomial distribution Compute P(43 X 51) when n = 6, p = 3/4 Binomial: n = 6, p = 3/4 k P(X = k) = ( ) 6 k (.75) k (.25) 6 k Total.7544 Mean µ = np = 6(3/4) = 45 Standard deviation σ= np(1 p) = 6(3/4)(1/4) = Mode (k with ma ) np + p = 6(3/4) + (3/4) = = 45 Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

12 Mode of a distribution The mode of random variable X is the value k at which the is maimum. Mode of binomial distribution when < p < 1 The mode is (n + 1)p. Eception: If (n + 1)p is an integer then (n + 1)p and (n + 1)p 1 are tied as the mode. The mode is within 1 of the mean np. When np is an integer, the mode equals the mean. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

13 Binomial and normal distributions Binomial Normal approimation to binomial k P(X = k).15 Binomial: n=6, p=3/4 Binomial P(43! X! 51) Normal: µ=45,"= Total P(X = k) shown as a rectangle: height P(X = k), etent k ± 1/2. The binomial distribution is only defined at the integers, and is very close to the normal distribution shown. We will approimate the probability P(43 X 51) we had above by the corresponding one for the normal distribution. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

14 Normal approimation to binomial, step 1 Compute corresponding parameters We want to approimate P(a X b) in a binomial distribution. We ll use n = 6, p = 3/4 and approimate P(43 X 51). Determine µ, σ: µ = np = 6(3/4) = 45 σ = np(1 p) = The normal distribution with those same values of µ, σ is a good approimation to the binomial distribution provided µ ± 3σ are both between and n. Check: µ 3σ 45 3(3.354) = µ + 3σ (3.354) = are both between and 6, so we may proceed. Note: Some applications are more strict and may require µ ± 5σ or more to be between and n. Since µ + 5σ , this would fail at that level of strictness. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

15 Normal approimation to binomial, step 2 Continuity correction Normal approimation to binomial Binomial: n=6, p=3/4 Binomial P(43! X! 51) Normal: µ=45,"=3.35 The binomial distribution is discrete (X = integers) but the normal distribution is continuous The sum P(X = 43) + + P(X = 51) has 9 terms, corresponding to the area of the 9 rectangles in the picture. The area under the normal distribution curve from 42.5 X 51.5 approimates the area of those rectangles. Change P(43 X 51) to P(42.5 X 51.5). Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

16 Normal approimation to binomial, steps Convert to z-scores 4. Use the normal distribution to approimately evaluate it For random variable X with mean µ and standard deviation σ, The z-score of a value is z = E(X) SD(X) The random variable Z is Z = X E(X) SD(X) Convert to z-scores: P(42.5 X 51.5) = P = µ σ. = X µ σ. ( X ) = P( Z ) Approimate this by the standard normal distribution cdf: Φ( ) Φ( ) This is close to the true answer (apart from rounding errors) P(43 X 51) =.7544 we got from the binomial distribution. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

17 Estimating fraction of successes instead of number of successes What is the value of p in the binomial distribution? Estimate it: flip a coin n times and divide the # heads by n. Let X = binomial distribution for n flips, probability p of heads. Let X = X/n be the fraction of flips that are heads. X is discrete, with possible values n, 1 n, 2 n,..., n n {(. n ) P(X = k n ) = P(X = k) = k p k (1 p) n k for k =, 1,..., n; otherwise. Mean E(X) = E(X/n) = E(X)/n = np/n = p. Variance Var(X) = Var ( ) X n = Var(X) n 2 Standard deviation SD(X) = p(1 p)/n. = np(1 p) n 2 = p(1 p) n. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

18 Normal approimation for fraction of successes n flips, probability p of heads, X=observed fraction of heads Mean E(X) = p Variance Var(X) = p(1 p)/n Standard deviation SD(X) = p(1 p)/n The Z transformation of X is Z = X E(X) SD(X) = X p p(1 p)/n and value X = has z-score z = p p(1 p)/n. For k heads in n flips, The z-score of X = k is z 1 = k np. np(1 p) The z-score of X = k/n is z 2 = (k/n) p p(1 p)/n. These are equal! Divide the numerator and denominator of z 1 by n to get z 2. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

19 Normal approimation for fraction of successes For n = 6 flips of a coin with p = 3 4, we ll estimate P ( 43 6 The eact answer equals P(43 X 51) X 51 6). Step 1: Determine mean and SD E(X) = p =.75 SD(X) = p(1 p)/n = (.75)(.25)/6 = Verify approimation is valid: Mean ± 3 SD between and 1 Mean 3 SD = Mean + 3 SD =.9177 Both are between and 1. Step 2: Continuity correction P ( ) ( X 6 = P Step 3: z-scores X Step 4: Evaluate approimate answer using normal distribution Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36 )

20 Normal approimation for fraction of successes P ( 43 6 ) ( 51 X 6 = P X = P(.7833 X.85833) (.7833 E(X) = P SD(X) ) X E(X) SD(X) = P ( Z = P( Z ) E(X) SD(X) ) ) = Φ( ) Φ(.74535) Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

21 Mean and SD of sums and averages of i.i.d. random variables Let X 1,..., X n be n i.i.d. (independent identically distributed) random variables, each with mean µ and standard deviation σ. Let S n = X X n be their sum and X n = (X X n )/n = S n /n be their average. Means: Sum: E(S n ) = E(X 1 ) + + E(X n ) = n E(X 1 ) = nµ Avg: E(X n ) = E(S n /n) = nµ/n = µ Variance: Sum: Var(S n ) = Var(X 1 ) + + Var(X n ) = n Var(X 1 ) = nσ 2 Avg: Var(X n ) = Var(S n )/n 2 = nσ 2 /n 2 = σ 2 /n Standard deviation: Sum: SD(S n ) = σ n Avg: SD(X n ) = σ/ n Terminology for different types of standard deviation The standard deviation (SD) of a trial (each X i ) is σ The standard error (SE) of the sum is σ n The standard error (SE) of the average is σ/ n Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

22 Theorem (Central Limit Theorem abbreviated CLT) For n i.i.d. random variables X 1,..., X n with sum S n = X X n and average X n = S n /n, and any real numbers a < b, P ( a S n nµ σ n ) b = P ( a X ) n µ σ/ n b Φ(b) Φ(a) if n is large enough. As n, the approimation becomes eact equality. 1.5 Binomial n=1,p=.75; µ=.75,!= Binomial n=6,p=.75; µ=45.,!= Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

23 Interpretation of Central Limit Theorem As n increases, the more and more closely resembles a normal distribution. However, the is defined as in-between the red points shown, if it s a discrete distribution. The cdfs are approimately equal everywhere on the continuum. Probabilities of intervals for sums or averages of enough i.i.d. variables can be approimately evaluated using the normal distribution. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

24 Repeated rolls of a die One roll: µ = 3.5, σ = 35/ Average of 1 roll of die; µ=3.5,!= Average of 2 rolls of die; µ=3.5,!= Average of 3 rolls of die; µ=3.5,!= Average of 1 rolls of die; µ=3.5,!=.17 Die average Normal dist. µ µ±! Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

25 Repeated rolls of a die Find n so that at least 95% of the time, the average of n rolls of a die is between 3 and 4. ( ) P(3 X 4) = P 3 µ σ/ X µ n σ/ 4 µ n σ/ n Plug in µ = 3.5 and σ = 35/12. ( P(3 X 4) = P 1/2 Z 35/(12n) ) 1/2 35/(12n) Recall the center 95% of the area on the standard normal curve is between z = ± / n (1.96) 2 35/ /(12n) (1/2) 2 n is an integer so n 45 Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

26 Sawtooth distribution (made up as demo) One trial: µ = 4, σ 2.24 Average of 1 trial; µ=4.,!=2.24 Average of 2 trials; µ=4.,!= Average of 3 trials; µ=4.,!= Average of 1 trials; µ=4.,!= Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

27 Binomial distribution (n, p) A Bernoulli trial is to flip a coin once and count the number of heads, { 1 probability p; X 1 = probability 1 p. Mean E(X 1 ) = p, standard deviation SD(X 1 ) = p(1 p). The binomial distribution is the sum of n i.i.d. Bernoulli trials. Mean µ = np, standard deviation σ = np(1 p). The binomial distribution is approimated pretty well by the normal distribution when µ ± 3σ are between and n. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

28 Binomial distribution (n, p) One flip: µ = p =.75, σ = p(1 p) = Binomial n=1,p=.75; µ=.75,!= Binomial n=6,p=.75; µ=4.5,!= Binomial n=3,p=.75; µ=22.5,!= Binomial n=6,p=.75; µ=45.,!= Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

29 Poisson distribution (µ or µ = λ d) Mean: µ (same as the Poisson parameter) Standard deviation: σ = µ. It is approimated pretty well by the normal distribution when µ 5. The reason the Central Limit Theorem applies is that a Poisson distribution with parameter µ equals the sum of n i.i.d. Poissons with parameter µ/n. The Poisson distribution has infinite range =, 1, 2,... and the normal distribution has infinite range < < (reals). Both are truncated in the plots. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

30 Poisson distribution (µ) Poisson µ=1;!= Poisson µ=6;!= Poisson µ=3;!= Poisson µ=6;!= Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

31 Geometric and negative binomial distributions Geometric distribution (p) X is the number of flips { until the first heads, (1 p) 1 p if = 1, 2, 3,... ; p X () = otherwise. The plot doesn t resemble the normal distribution at all. Mean: µ = 1/p Negative binomial distribution (r, p) Standard deviation: σ = 1 p/p r = 1 is same as geometric distribution. r > 2: The has a bell -like shape, but is not close to the normal distribution unless r is very large. Mean: µ = r/p Standard deviation: σ = r(1 p)/p Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

32 Geometric and negative binomial distributions Heads with probability p =.1 Geometric p=.1; µ=1.,!= Neg. bin. r=6,p=.1; µ=6.,!= Neg. bin. r=3,p=.1; µ=3.,!= Neg. bin. r=6,p=.1; µ=6.,!= ! Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

33 Eponential and gamma distributions Eponential distribution (λ) The eponential distribution doesn t resemble the normal distribution at all. Mean: µ = 1/λ Standard deviation: σ = 1/λ Gamma distribution (r, λ) The gamma distribution for r = 1 is the eponential distribution. The gamma distribution for r > 1 does have a bell -like shape, but it is not close to the normal distribution until r is very large. There is a generalization to allow r to be real numbers, not just integers. Mean: µ = r/λ Standard deviation: σ = r/λ Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

34 Eponential and gamma distributions Rate λ =.1 Eponential!=.1; µ=1.,"= Gamma r=6,p=.1; µ=6.,!= Gamma r=3,p=.1; µ=3.,!= ! Gamma r=6,p=.1; µ=6.,!= ! Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

35 Geometric/Negative binomial vs. Eponential/Gamma p = λ gives same means for geometric and eponential. p = 1 e λ gives same eponential decay rate for both geometric and eponential distributions. 1 e λ λ when λ is small. This corespondence carries over to the gamma and negative binomial distributions. Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

36 Geometric/negative binomial vs. Eponential/gamma This is for p =.1 vs. λ =.1; a better fit for λ =.1 would be p = 1 e λ.95 Geometric p=.1; µ=1.,!=9.49 Eponential!=.1; µ=1.,"= Neg. bin. r=3,p=.1; µ=3.,!= Gamma r=3,p=.1; µ=3.,!= ! Prof. Tesler , 2.1 Normal distribution Math 283 / October 22, / 36

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