Mathacle. PSet Stats, Concepts In Statistics Level Number Name: Date: , the DeMoivre-Laplace Theorem states: ( ) n x.

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1 6.4. Approimtion of Binomil Distriution y the Norml Distriution n n [MATH] When np ( p) >>, y using the Stirling s formul n! n e πn, the DeMoivre-Lplce Theorem sttes: ( ) ( n, p, ) = C p p n n e π np( p) ( np) np( p) = e np( p) π np np( p) = e σ π µ σ z where σ = Vr( ) = np( p), µ = np. Let φ( z) = e, the stndrd norml π distriution function or the Z-distriution function: µ ( n, p, ) φ σ σ 84

2 µ If z =, which is known s the z-scores, the inomil distriution cn e σ pproimted y the Z-distriution ( n, p, ) φ ( z) = φ ( z) z σ Where z = cn e viewed s the increments. σ In prctice, the conditions np > 0 nd n ( p) > 0 re good pproimtion for the conditions np ( p) >>. Let q = p, nd rememer tht 0 q, p. The 99% of re of the distriution is symmetriclly spnned within 6 stndrd devitions, so µ > 3σ is good pproimtion. Tht is, np > 3 npq or n p > npq np > 9q So, select np > 0 should e sufficient to stisfy the ove inequlity. Similrly, So nq > 0 is the other condition. nq > npq > 0q 85

3 Pr when the conditions re µ stisfied, where nd re different non-negtive integers, let z = nd σ µ z = : σ [MATH] To pproimte the inomil proility ( X ) z ( X ) = ( n p ) Pr,, = µ φ d σ σ µ z= σ z z ( z ) φ( z) dz = Φ Φ( z ) = Pr( z < z < z ) Where Φ ( z) = φ( z) dz, the CDF of the Z-distriution. The pproimtion of the integrl cn e viewed s to find the re under the curve with increments µ =, z = =. Since nd re integers, ± or ± my e used σ σ for etter pproimtion of the ounds when it comes to different cses:.) = =, use, +.), use s the lower ound. 86

4 c.), use + s the upper ound. d.), use, + s the ounds. Emple For the following given inomil distriutions, use the inomil formul nd norml formul to clculte the proilities when.) n = 0, p = 0. 5, X = 3, np =, nq = Binomil: Norml:.) n = 0, p = 0. 5, 4 X 7, np =, nq = Binomil: Norml: c.) n = 00, p = 0. 5, X = 30, np =, nq = Binomil: 87

5 Norml: d.) n = 00, p = 0. 5, 40 X 70, np =, nq = Binomil: Norml: Emple Estimte the proility of ectly 55 heds in 00 tosses of coin. Given: n = 00, = 55, p = 0. 5, q = 0. 5 The conditions for using norml pproimtion: np = 50 >> 0, nd n ( p) = 50 >> 0 The men nd stndrd devition: µ = np = = 50, σ = np( p) = ( 0.5) = (00, 0.5,55) φ( ) = φ() = e π 88

6 Emple You re plnning smple survey of smll usinesses in your re. You will choose n SRS of usinesses listed in the telephone ook s Yellow Pges. Eperience shows tht only out hlf the usinesses you contct will respond. Suppose tht you contct 50 usinesses..) Is it resonle to use the inomil distriutions or norml distriutions to study the numer rndom vrile X s the numer of responses?.) Wht is the proility tht 70 or fewer will respond? (Use oth inomil nd norml distriutions, if they re pplicle).) It is resonle to ssume s inomil ecuse B: Respond or not respond I: Assume the clls re independent N: n = 50 S: p = 0.5, q = 0.5 It is resonle to use norml distriution ecuse np = 50(0.5) = 75 0, n( p) = 50( 0.5) = 75 0 The men is µ = np = 50(0.5) = 75, nd stndrd devition is σ npq = 50(0.5)(0.5) = 6.4 = = 0.5, = = 70.5.) ( ) P 0 X 70 inomcdf (75, 0.5, 70) P ( 0 X 70) Φ Φ

7 Emple Are ttitudes towrd shopping chnging? Smple surveys show tht fewer people enjoy shopping thn in the pst. A recent survey sked ntionwide rndom smple of 500 dults if they greed or disgreed tht I like uying new cloths, ut shopping is often frustrting nd time-consuming. The popultion tht the poll wnts to drw conclusions out is ll U.S. residents ged 8 or over. Suppose tht in fct 60% of ll dults U.S. residents would sy Agree if sked the sme question..) Wht is the proility tht 50 or more of the smple gree?.) Wht is the proility tht t most 468 people in the smple would gree? n = 500, = 50, = 468, p = 0.6, np > 0, n( p) > 0, n < 0% N, µ = np = 500(0.6) = 500, σ = npq = 500(0.6)(0.4) = = = 59.5, = = ) z µ = = = 0.796, Φ( Z z ) = σ Direct inomil clcultion is P( X 50) = P( X 59) = inomcdf (500, 0.6,59) = = 0.3 µ ) z = = =.859, Φ( Z z ) = Direct inomil σ clcultion is P( X 468) = inomcdf (500,0.6,468) =

8 Modeling Discrete Distriution with Norml Distriution Binomil distriution cn e pproimted y the norml distriution when np > 0 nd n ( p) > 0. Another cse is when the smple size is lrge enough (>30), the smple mens re pproimted y the norml distriution y the Centrl Limit Theorem, which will e discussed lter. The emple elow illustrtes how distriution cn e curvefitted in norml distriution. The following histogrm grph displys 0,000 rndom digits of,, 3,, 9 generted from specific rndom digit genertor. Assume tht smple with size of 0000 ws lrge enough, so the histogrm reflects the true distriution of this rndom digit genertor. Curve-fit the distriution into the norml distriution model is sed on the fcts tht 99% of the dt should e covered within the rnge of 6σ nd the men is locted t the center. Since the mode is t 4 nd 99% of digits lie within to 7 inclusive, then the men nd 7 stndrd devition cn e pproimted y µ = 4, 6σ = 7 σ =, p( ) = e. Some of dt otined from oth distriution nd the model. 7 π distriution model Think pro p( ) = p( ) under the norml distriution, where =. The model seems resonle. The model is more ccurte towrd the men thn towrd the tils. More dvnced techniques eyond this notes cn e used to verify the ccurcy of the modeling. Norml distriution is widely used, or used, in mny pplictions. A good prctice is to t lest understnd why nd how norml distriution is implemented. 9

9 Quick-Check Approimtion of Binomil Distriution By Norml Distriution QC Suppose smple of 600 tires of the sme type is otined t rndom from n ongoing production process in which 8% of ll such tires produced re defective. Wht is the proility tht in such smple50 or fewer tires will e defective? QC Premture ies re those orn efore 37 weeks, nd those orn efore 34 weeks re most t risk. It ws reported tht % of irths in US occur efore 34 weeks. If,000 irths re selected rndomly, wht is the proility there re 0 to 5 ies, inclusive, re t risk? 9

10 Answers QC Given: n = 600, = 50, p = 0. 08, 0 X 50, = = 0. 5, = = The conditions for using norml pproimtion: np = = 8 >> 0, nd n ( p) = = 47 >> 0 The men nd stndrd devition: QC µ = np = = 8, σ = np( p) = ( 0.08) = P ( 0 X 50) = Φ Φ n = 000, = 0, = 5, p = 0.0, = = 9.5, = = 5.5 np = 000(0.0) > 0, nq = 000(0.98) > 0, µ = np = 000(0.0) = 0, µ σ = npq = 000(0.0)(0.98) = 4.47, z = = =.37, σ 4.47 µ z = = =.4, P( z Z z ) = σ

11 6.5. The Norml Distriution nd Its Z-Scores The norml distriutions, commonly known ell curve, re derived from the inomil distriutions under certin conditions. The curve is only chrcterized y the men nd stndrd devition. A specil cse is when the men is zero nd stndrd devition is one. This distriution is clled the Stndrd Norml Distriution, or Z-Distriution. Other norml distriutions cn e trnsformed into the stndrd norml distriution y using Z- scores trnsformtion. Stndrd Norml Distriution or Z Distriution N (0,) The stndrd norml proility density function (pdf) of rndom vrile Z is given y (z) φ : z φ(z) = e π The distriution hs men of zero nd stndrd devition of one. It is denoted s Z ~ N(0,). 94

12 [MATH] Proility for the Z-distriution is the re under the φ(z) curve: z z z Pr( z Z z ) = φ(z) dz = e dz z z π In terms of CDF function Φ ( z) : z Where ( z) φ( z) dz Φ =. Pr( z Z z ) = Φ( z ) Φ ( z ) [Ti-84] The re under the curve φ (z) for given intervl z, z ) ( Pr( z < z < z ) = normlcdf ( z, z,0,) cn e clculted y The vrile z vlue for given re A = Pr( < z < z ) cn e clculted y z = invnorm(a, 0,) Note tht Pr( < z < + ) =, Tht is, the entire re under the curve φ (z) is one. For the clcultion purposes, when z or z re infinity, lrge numer such s 000 cn e used in the plce of infinity. Emple Given the Z-distriution N (0,), solve ech of the prolems elow. Lel the domin (intervl) nd shde the proility re..) Pr( < z < ).) Pr( z < ) 95

13 3.) Pr( z < ) 4.) Pr( < z < ) 5.) Find z, where Pr( < z < z ) = 0.05, or the proility tht is elow 5 th percentile, 6.) Find z, where Pr( z < z < + ) = 0., or the proility tht is ove 80 th percentile. 96

14 7.) Find z *, where Pr( z < z*) = ) Find z *, where Pr( z > z*) = ) Find the integer n such tht.96 0 Pr z < = n 0.) Find σ such tht.64σ Pr( z < ) = ) Pr( < z < ) = normlcdf (,, 0, ) = ) Pr( z < ) = AREA( < z < )

15 3.) Pr( z < ) = AREA(0 < z < ) ) The cn e pproimted y reltive lrge numer such s 00. Therefore, Pr( < z < ) Pr( 00 < z < ) = normlcdf ( 00,, 0, ) = ) z = Since Pr( < z < z ) = 0.05, z = invnorm( 0.05, 0, ) = ) z = The proility on the left side is 0. = 0. 8, z = invnorm( 0.8,0,) = ) z * =. 96. Since z * < z < z *, z = z * nd z = z *. The proility on the left 0.05 side is = , z* = z = invnorm(0.975,0,) = ) z * =. 64. Since z * > z or z > z *. The proility on the left side for clculting 0. the z * is = 0. 95, z * = invnorm(0.95,0,) =

16 ) n = 400. Since =.96 n = 400 n.64σ 0.) σ =0 0. Since =.64 σ = Norml Distriution (, ) N µ σ nd Z-Scores A continuous rndom vrile X hs norml distriution with prmeter µ nd σ, denoted y N ( µ, σ ), if the proility density function is given y µ σ p( ) = e, for R σ π Where µ = E[X ] nd σ = Vr[X ]. Tht is, the men, vrince nd stndrd devition of the norml distriutions re µ, σ ndσ, respectively. The norml density function p( ) stisfies the following properties: p( ) is symmetric to = µ ; p( ) increses, nd then decreses with mode = µ ; p( ) chnges concvity t = µ ± σ. Pr( X ) = p() d = F( ) F( ), where F( ) = p() d is the CDF. 99

17 The function φ (z) is ssocited with function p () through the liner trnsformtion: µ z = σ [MATH] The functions nd p nd φ re connected in following formul: µ p( ) = φ = φ ( z) σ σ σ The z-scores re the normlized rndom vrile with zero s the men. The proility µ µ Pr( X ) = F( ) F( ) = Φ Φ = Φ z Φ z σ σ Or simply, Pr( X ) = Pr( z Z z ) ( ) ( ) [Ti-84] Clculte the proility of rndom vrile with norml-distriution N ( µ, σ ) for given intervl (, ) : ( ) Pr( X ) = p() d = normlcdf,, µ, σ When F( ) = p( ) d is given, the upper ound cn e found y ( ( ),, ) = invnorm F µ σ 00

18 The proility under φ (z) in the intervl I z is the sme s the proility under p() in I, where I [ z, z ] = nd I [, ] [ µ z σ, µ z σ ] z = = + +. Emple Given the proility under the φ (z) is Pr( < z < z ) = 0.95, µ = 0 nd σ =, find, nd Pr( < < ) for the proility under p( )? From the given, z =, = invnorm( 0.95, 0, ) = Becuse z z = µ, σ 3. = µ + z σ = 0 + ( )() =, nd = µ + z * σ = () = 9. Since z =, choose lrge numer such s 000 s pproimtion for z. Then, Pr( < < 3.9) = normlcdf ( 000, 3.9, 0, ) Emple Given Pr( z < z*) = 0.95, µ = 0 nd σ =, find nd. Since z * < z < z *, this implies z = z *, z = z *, nd Pr( < z < z*) = The z * = invnorm(0.975, 0, ) =.96, = µ + ( z*) σ = 0.96() = 6.08, = µ + z * σ = () = 3.9, nd I z = [.96,.96], I = [6.08, 3.93]. The proility Pr(6.08 < < 3.93) = normlcdf (6.08, 3.95, 0, ) Emple The verge nnul rinfll t El Yunque Rinforest on the islnd of Puerto Rico is 00 inches. Assume the vrince to e 5 inches nd the distriution is pproimtely norml. Wht is the proility tht the rinfll there net yer will e etween 85 nd 30 inches? 0

19 Pr(85 X 30) = F(30) F(85) = normlcdf (85, 30, 00, 5) Or, use the z-scores, z85 =.34, z30 =.68, nd the proility is 5 5 Pr(85 30) = Pr(.34 < z <.68) = normlcdf (.34,.68, 0.) Emple The verge Mth SAT score for grduting seniors t THS is 640 with stndrd devition of 30. In terms of SAT mth score, wht Mth SAT score would e needed to e in the top 0% of grduting senior students? Assume the scores re pproimte normlly distriuted. µ = 640 nd σ = 30. The z-score for 90 th percentile is z = invnnorm(0.9,0,).8. So, the scores tht re greter thn 90 th percentile is > µ + zσ = (.8) = 678. Emple At THS the men numer of colleges tht seniors pply to is 5.6 with stndrd devition of.3. Assuming the distriution is pproimtely norml, wht percent of seniors t THS pply to more thn 8 schools? µ = 5.6 nd σ =.3. The z-score for equls to 8 is z =.0434 nd.3 proility Pr( > 8) = Pr( z >.0434) = normlcdf (.0434,000, 0,0) 4.8%. Emple The verge time it tkes for men to run mrthon is 4 hours nd 4 minutes with stndrd devition of hour nd minutes. The verge time it tkes for women to run mrthon is 4 hours nd 5 minutes with stndrd devition of hour nd 7 minutes. Ptrick Mku of Keny holds the world record for men with hours nd 3 minutes nd Pul Rdcliffe of UK holds the world record for women with hours nd 0

20 5 minutes. Reltive to other mrthoners of like gender, who hs etter world record time? Justify your nswer sttisticlly. For Ptrick, For Pul, z ptrick z pul Pul hs etter time. 3/ / 60 =.7, or.7σ elow the men. 6 / / 60 9 / 60 =.34, or.34σ elow the men. 67 / 60 Emple Hens usully egin lying eggs when they re out 6 months old. Young hens tend to ly smller eggs, often weighing less thn the desired minimum weight of 54 grms..) The verge weight of the eggs produced y the young hens is 50.9 grms, nd only 8% of their eggs eceed the desired minimum weight. If norml model is pproprite, wht would the stndrd devition of the eggs weights e?.) By the time these hens hve reched the ge of yer, the eggs tht they produce verge 67. grms, nd 98% of them re ove the minimum weight. Wht is the stndrd devition for the pproprite norml model for these older hens? c.) A certin poultry frmer finds tht 8% of his eggs re underweight nd tht % weigh over 70 grms. Estimte the men nd stndrd devition of his eggs. 03

21 µ ). µ = 50.9, = 54, only 8% ove 54, so z = 0.588, σ = = = 5.38 z µ ). µ = 54, = 67, only 98% ove 54, so z =.05, σ = = = 6.39 z.05 c.). = 54, the underweight, = 70, z =.405, z =.749, µ 54 µ µ 70 µ = =.405, = =.749 σ σ σ σ µ = 6.7, σ = 6. Emple [CB] Some descriptive sttistics for set of test scores re shown elow. For this test, certin student hs stndrdized score of z =.. Wht score did this student receive on the test? =.9 = Emple [MC4] A distriution of scores is pproimtely norml with men of 78 nd stndrd devition of 8.6. Which of the following equtions cn e used to find the score ove which 33 percent of the scores fll? 04

22 The nswer is D. The criticl z vlue for 33% of the re is ( z * = invnorm( 33%) = ) Emple [FRQ0M] A chrity fundriser hs Spin the Pointer gme tht uses fund rising. Bsed on lst yer s event, the chrity nticiptes tht the Spin the Pointer gme will e plyed 000 times. The chrity would like to know the proility of otining net contriution of t lest $500 in 000 plys of the gme. The men nd stndrd devition of the net contriution to the chrity in 000 plys of the gme re $700 nd $9.79, respectively. Use the norml distriution to pproimte the proility tht the chrity would otin net contriution of t lest $500 in 000 plys of the gme. µ = 700, σ = 9.79, = z = = P( Z >.55) = 98.44% 05

23 Quick-Check The Norml Distriution nd Z Scores QC Given the Z-distriution N (0,), solve ech of the prolems elow. Lel the domin (intervl) nd shde the proility re. Round your nswers to three digits..) Proility: Pr( Z 3) =.) Proility: 97.5 th percentile or Pr( Z z*) = 0.05, z* = c.) Proility: the middle 95% or Pr( Z z*) = 0.95, z* = QC The heights of mle students t THS re pproimtely norml with men of 68 inches nd stndrd devition of.5 inches. Aout 0% of these mle students re tller thn wht height? 06

24 QC Jenn tkes three SAT suject tests. She scores 600 on ll three. Using stndrd scores, or z-scores, rnk her performnce on the three tests from the est to worst if mens nd stndrd devitions for the tests re s follows: Men Stndrd Devition Test I Test II Test III QC A student in Science Reserch clss interested in the ge t which women re hving their first child surveyed simple rndom smple (SRS) of 50 women hving t lest one child nd found n pproimtely norml distriution with men ge of 7.3 nd stndrd devition of 5.4. Approimtely the middle of 95% of women hd their first child etween wht ges? QC Luren scored n 8 on Wissler s geometry test for which the clss men ws 74 with the stndrd devition of 3.. She scored n 86 on Nelson s iology test for which the clss men ws 77 with stndrd devition of.9. In comprison to other memers of ech of these clsses, in which clss did Luren perform etter? QC To get into Science Reserch clss, freshmen t THS require to score in the upper 0% on mth plcement test. If the men score on the test is 50 nd the stndrd devition of the scores is 80, wht is the minimum score tht student cn ern on the test to meet the dmission requirement? Scores on the test re normlly distriuted nd re reported in the intervls of 0. 07

25 QC The GPAs re normlly distriuted with stndrd devition of 0.7. Wht is the men GPA if 5% of the students hve GPA ove 4.75? QC A survey ws tken to determine the numer of hours per week students studied mth nd sciences. The distriution ws pproimtely norml with men of 0 hours nd stndrd devition of 4 hours. Approimtely wht percent of the students studied etween 7 nd 30 hours? QC Grdes on sttistics test re normlly distriuted with men of 76 nd stndrd devition of 8. The top 0% students will receive n A, the net 0% B, the net 40% C, the net 0% D, nd the ottom 0 n F. The students who received F s in course hd no higher thn wht score? QC The men score on Trig & Stts test ws 75 with stndrd devition of four. Wht would e the z-score for Alejndro who scored n 8? QC John needs to hve tree cut down in his ckyrd. A tree service gives him n estimte of $6500. John feels this is too high nd clls the Better Business Bureu which reports the verge cost for tht service in his re is $400 with stndrd devition of $750. Assuming tht the costs re normlly distriuted, John concludes tht pproimtely 95% of the time, the cost should roughly fll etween which two vlues? 08

26 Answers QC ) 0.997,.).96, c.) z*=.96 QC in. µ = 68, σ = 8.5, z* = invnorm(0.8, 0.) = 0.34, 68 > 0.84 = in Pr > 0.84 = 0., 8.5 QC III>I>II. The higher the z-scores, the etter she did in the test. z I = =.5, 80 z II = =.08, 0 z I = = QC (6.7, 37.88) QC She did etter in Biology clss. QC QC QC QC QC QC (730, 5670) 09