Stat 225 Lecture Notes Week 7, Chapter 8 and 11

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Normal Distributio Stat 5 Lecture Notes Week 7, Chapter 8 ad Please also prit out the ormal radom variable table from the Stat 5 homepage. The ormal distributio is by far the most importat distributio that we will study. It is the distributio that is most used i daily life to model real world situatios. It was first used bi 733 by Abraham DeMoive ad used Carl Friedrich Gauss to predict the locatio of astroomical bodies. We call it ormal because it is applicable to most ormal pheomea for a large populatio/ Situatio: Much data arises from repeated measuremets o the same variable. Examples Test scores for all Stat 5 courses. Height of 000 male football players. Sales receipts for 000 customers at a grocery store. PDF: A cotiuous radom variable is said to have a Normal distributio with mea μ ad variace >0, if it had the PDF x μ f ( x) = e for < x < π Note: A ormal distributios ca be completely determied by the mea ad stadard deviatio. Notatio: or ~ Normal( μ, ) ~ N( μ, ) Shape:

Expectatio ad Variace E ( ) = μ Var( ) = Characteristics: This distributio is ofte also referred to as the Gaussia distributio. The PDF has the characteristic bell shape ad symmetric about the mea (higher desity i the middle tha the tails). Sice this distributio is symmetric it has the property that the mea is equal to the media. The mea μ determies the locatio of the curve, ad determies the shape (scale or sharpess) of the curve (see the attached graph). Area uder the curve from egative ifiity to x is P( < x) = F(x). Fact: Liear combiatios of idepedet Normal radom variables are themselves Normal. Let ~ Normal( μ, ) ad ~ Normal( μ, ) ad a, b, c costats, the a + b ~ Normal( aμ + b, a ) a + b + c ~ Normal( aμ + bμ + c, a + b ) Stadard Normal Distributio: A ormal radom variable with mea μ = 0 ad variace = is called a Stadard Normal Radom Variable. Notatio: Z ~ Normal( μ = 0, = ) Commet: Ay Normal( μ, ) radom variable ca be coverted to a Stadard Normal radom variable. We refer to this process is called stadardizatio. Corollary: Process of Stadardizatio. If ~ Normal( μ, ), the what should a ad b be so that Z = a + b ~ Normal(0,)? a = b = μ μ Z=

Empirical Rule for a Stadard Normal Variable Withi stadard deviatio there is 68% of the data. Withi stadard deviatios there is 95% of the data. Withi 3 stadard deviatios there is 99.7% of the data. Computig Probabilities How do you compute the probabilities for a Normal radom variable? The PDF is ot easy to itegrate ad there does ot exist a closed form of the CDF for geeral μ ad, but the CDF values of a stadard ormal radom variable ca be foud i tables (ad with may calculators ). CDF: The otatio used for the CDF of a stadard Normal distributio is Φ( z) = P( Z z). You ca use Excel (commad: NORMDIST(x, mea, stadard deviatio) ) to compute CDF values for geeral ormal radom variables. Commets: If we wat to compute probabilities for ~ Normal( μ, ) (without usig a computer), the we first have to stadardize ito μ Z = ad the read the values from the stadard Normal table. 3

Examples. Fid the followig probabilities give Z ~ Normal( μ = 0, = ) a) P( Z.75) b) P( Z ) i. Usig the empirical rule ii. Usig the table c) P( Z >.) d) P(. Z ) e) What is a if P(Z > a) = 0.44? f) What is a if P( a Z a) = 0.4?. For ~ Normal( μ =, = 4) a) Fid P( < ) b) Fid P( < < ) c) What is b so that P( < b) = 0.08? d) The middle 90% observatios of fall betwee which two values? 4

e) Fid P ( >.44) f) What is P(3 + <5.5)? g) What is distributio of Y = 3 +? 3. Let ~ Normal( μ =, = ) ad ~ Normal( μ = 3, = ) be idepedet. Fid: c) P + ) ( b) P( + > ) The table lists values of the stadard Normal CDF, deoted Φ (z) for may differet values of z. Give z, you ca read off the CDF value Φ (z). But the table works i reverse, too. 3. For ~ Normal( μ =, = 3) fid x, such that P ( x) = 0.0. 5

4. I a study of elite distace ruers, the mea weight was reported to be 63. kg with a stadard deviatio of 4.8. Assumig the distributio of the weights is ormal, sketch the desity curve of the weight distributio, with horizotal axis marked first i stadard deviatios, ad below marked i kgs. a) Fid the percet of ruers whose weight was greater tha 69 kg. b) Te elite ruers are weighted oe by oe, what is the probability that at least of them exceed 69 kg? Practice Problems. Usig the table, fid the proportio of observatios from a stadard ormal distributio that satisfies each of the followig statemets: a) For Z ~ Normal ( μ = 0, = ) fid P(Z = 0.5) b) For Z ~ Normal ( μ = 0, = ) fid P (.5 < Z < 0.8) c) For Z ~ Normal ( μ = 0, = ) fid P( Z >.5) d) For ~Normal ( μ =, = 4 ) fid P( > ) 6

e) For ~ Normal ( μ = 3, = 4 ) fid P( < ). Let ~Normal ( μ = 0, = ) ad ~Normal ( μ =, = 4 ) a) Fid P( - 3 > 7) b) Fid P( - > ) 3. The graduate Record Examiatios are widely used to help predict the performace of applicats to graduate schools. The rage of possible scores o a GRE is 00 to 900. The psychology departmet at a uiversity fids that the scores of its applicats o the quatitative GRE are approximately ormal with mea = 544 ad stadard deviatio = 03. Use the table to fid the relative frequecy (percet) of applicats whose score is a) less tha 500 b) betwee 500 ad 700. c) What miimum score would a studet eed i order to score better tha 77% of those takig the test? 7

4. A college exam for a course i the history of alie ecouters has a mea of 70 ad a stadard deviatio of 0. The professor determies that ay studet who scores i the bottom 0 th percetile will fail the course. a) Determie the scores o the exam that will produce a failig grade. b) Give a studet s score is greater tha 80, what is the probability that his core is less tha 90 5. A luch stad i the busiess district has a mea daily gross icome of $40 with a stadard deviatio of $50. Assume that the daily gross icomes are ormally distributed. What is the relative frequecy (percet) correspodig to a daily gross icome of $495 or more? 8

Normal Approximatio to the Biomial Whe is the ormal approximatio is used? Assume I toss 00 fair cois for 0000 times. Each time I record the umber of heads. Totally there are 0,000 records. If these 0,000 are put ito oe histogram, we will see that the distributio looks like the ormal distributio (bell-shaped ad symmetric). I aother word, if is the umber of heads amog 00, the follows ormal distributio approximately. Situatio: For a radom variable ~Biomial(, p) as, the PMF of the Biomial becomes very similar i shape to the Normal PDF. Due to this similarly we ca use a Normal radom variable to approximate a biomial radom variable if is large eough ad p reasoably close to 0.5. The followig graphs show the Biomial PMF of for p = 0.5 ad selected values of. Superimposed is the cotiuous curve of a Normal PDF. 9

As you ca see above, for large values of ad moderate (close to 0.5) values of p i the PMF os is very close to a Normal PDF with μ = p ad = p( p). Approximatio Criteria We will use the criteria p > 5 ad ( p) > 5 as a rule of thumb for whe to use a Normal Approximatio for a Biomial radom ~ Biomial (, p) variable with ~ Normal( μ = p, = p( p)). The larger is, the more precise this approximatio will be. Cotiuity Correctio There is oe importat detail that we have to take care of whe computig Normal approximatio of Biomial radom variables. ~ Biomial (, p) is a discrete radom variable where as a Normal radom variable is a cotiuous radom variable. For a Biomial radom variable there is a differece betwee P ( < x) ad P( x) where for a Normal radom variable there is o differece i these two terms. This meas that we have to either iclude or exclude the a -bar i our computatio of the probability. We ca do that be goig a step of ½ to the left or right depedig o whether we wat to iclude or exclude the bar. This procedure is called a Cotiuity Correctio. 0

Cotiuity Correctio Formula For limits a ad b we have, for example, P ( a b + μ a μ b) Φ Φ The Cotiuity Correctio ca also be used to compute probabilities like P( a < b), but the the directio of the step eeds to be chaged. Example: It is kow that a machie has the probability 0. to make defective fuses. a) What is the probability that amog 000 such fuses, at least 60 ad at most 80 are defective oes? b) What is the probability that amog 000 such fuses, more tha 60 ad less tha 80 are defective oes? c) What is probability that over 80 fuses are defective? d) What is the probability that there are exactly 80 defective fuses?

Practice Problems Verify if the assumptios of the Normal approximatio are met for each of the problems below ad aswer the questio asked usig appropriate methods.. Accordig to govermet data, % of America childre uder the age of 6 live i households with icomes less tha the official poverty level. A study of learig i early childhood chooses a SRS of 300 childre. What is the probability that more tha 80 of the childre i the sample are from poverty households?.07. I a test of ESP (extrasesory perceptio), the experimeter looks at cards that are hidde from the subject. Each card is equally likely to cotai a star, a circle, a wavy lie, or a square. A experimeter looks at each of 00 cards i tur, ad the subject tries to read the experimeter's mid ad ame the shape o each. What is the probability that the subject gets 30 or more correct if the subject does ot have ESP ad is just guessig? (Assume the 00 observatios are idepedet.).49. The Purdue quarterback completes 44% of his passes. Fid the probability of the quarterback completig 5 or more of his ext 0 passes..005

Cetral Limit Theorem The Cetral Limit Theorem (CLT) is oe of the most importat theorems i probability theory. It was first stated by Abraham De Moivre ad geeralized by Pierre Laplace. Because of this, this theorem is also sometimes called the De Moivre Laplace Theorem. Situatio: Cosider a experimet where a certai variable is measured repeatedly may times. Let i be the result of the i th measuremet. The i s may have some discrete or cotiuous (ot ecessarily kow) distributio. What ca we say about the sum or average of the s? Demostratio: i Suppositio: If a measuremet is repeated may times, the the sum or average _ of the measuremets is _ approximately ormally _ distributed. 3

Defiitio: Let,..., be idepedetly ad idetically distributed (iid) radom variables. We defie the sample mea (average) to be Note: If ( ) = μ i = i= i E ad Var( ) = for all, the E( ) = E i = E i i= i= Var( ) = Var i = Var i ( ) Var = i= i= Cetral Limit Theorem: i ( ) = ( ) E( ) = E( ) = μ i ( ) ( ) i = Let,..., be idepedet radom variables with E ( i ) = μ ad Var( i ) = for i =,...,. The for large ( > 30 ), the sum ad the average of the i s has approximately a Normal distributio. Average: = Normal i ~ μ, i= Sum: i ~ Normal( μ, ) i= The larger is, the more closely will the PDF of the sum or average resemble a Normal PDF. If the i ~ N ( μ, ) to start with you do ot eed to meet the >30 criteria for the sum ad average to be ormally distributed. Examples. Oe has 00 light bulbs whose lifetimes are idepedet expoetials with mea 5 hours. If the bulbs are used oe at a time, with a failed bulb beig replaced immediately by a ew oe, what is the probability that there is a still workig bulb after 55 hours. 4

. How ofte do you have to roll a fair (six-sided) die, so that the probability that the average of the scores is betwee 3. ad 3.8 is at least 0.9? 3. Assume there are a average of 0.5 typos o each page of a ovel draft with 000 pages writte by a ovelist. Jim is a editor ad is goig to proofread the draft. What is the probability that for the first 50 pages, the average typos he foud o each page is betwee 0.44 ad 0.48? Practice Problems PP. Studet scores o exams give by a certai istructor have a mea 74 ad stadard deviatio 4. The istructor is about to give the exam to a class of size 40. Approximate the probability that the average test scores i the class of size of 40 exceeds 80. PP. A ew elevator i a large hotel is desiged to carry about 30 people, with a total weight of up to 5000 lbs. More tha 5000 lbs overloads the elevator. The average weight of guests at this hotel is 50 lbs with a SD of 55 lbs. Suppose 30 of the hotel's guests get ito the elevator. Assumig the weights of these guests are idepedet radom variables, what is the chace of overloadig the elevator? 5

PP3. The distributio of actual weights of 8-oz. chocolate bars produced by a certai machie is ormal with mea 8. ouces ad stadard deviatio 0. ouces. a) If a sample of five of these chocolate bars is selected, the probability that their average weight is less tha 8 oz. b) If a sample of five of these chocolate bars is selected, there is oly a 5% chace that the average weight of the sample of five of the chocolate bars will be below what average value Statistical Iferece: Cofidece Itervals Z-cofidece itervals for the populatio mea Why do we eve bother aalyzig data? We wat to draw coclusios from the data. Populatio: the etire group of idividuals that we wat iformatio about. Sample: a part of the populatio that we actually examie i order to gather iformatio about the whole populatio Parameter: a umber that describes the populatio a fixed umber, but i practice we do ot kow its value Note: The values of the parameters were give or you may figure them out i the previous chapters, so that you kow the exact distributio of the populatio ad you ca calculate the correspodig probabilities, etc. However, i this sectio, we will assume that they are ukow. Example: populatio mea μ. 6

Statistic: a umber that describes a sample its value is kow whe we have take a sample, but it ca chage from sample to sample (samplig variability) ofte used to estimate a ukow parameter +... + Example: sample mea =. Statistical Iferece: use a fact about a sample to estimate the truth about the whole populatio Why ca t we just accept our sample mea as the populatio mea μ? Every time whe we use the statistics (sample mea), we get a differet aswer due to samplig variability. Two most commo types of formal statistical iferece: Cofidece Itervals: whe we wat to estimate a populatio parameter Sigificace Tests: whe we wat to assess the evidece provided by the data i favor of some claim about the populatio (yes/o questio about the populatio) Cofidece Itervals allow us to estimate a rage of values for the populatio mea kow how cofidet we should be that the populatio mea is withi that rage The true mea for the populatio exists ad is a fixed umber, but we just do t kow what it is. Usig our sample statistic, we ca create a et to give us a estimate of where to expect the populatio parameter to be. If we just take a sigle sample, our sigle cofidece iterval et may or may ot iclude the populatio parameter. However if we take may samples of the same size ad create a cofidece iterval from each sample statistic, over the log ru 95% of our cofidece itervals will cotai the true populatio parameter, if we are usig a 95% cofidece level. 7

If you icrease the size of your sample (), you decrease the size of your et (or your margi of error). If you icrease your cofidece level, the you icrease the size of your et (or your margi of error). A smaller et is good because it gives you more iformatio. It is a smaller rage for where to expect your true populatio parameter. Cofidece itervals look like: estimate ± margi of error Cofidece Iterval for a Populatio Mea μ: ± z * x (sample mea) is the estimate of μ; is the populatio stadard deviatio; x is the stadard deviatio of ; x m= z * x is called margi of error; 8

z* is the value o the stadard ormal curve with area C betwee z* ad z*. z*.645.960.576 C 90% 95% 99% is our guess for μ, ad the margi of error guess is. C is the cofidece level. z * x shows how accurate we believe our Remember that the mea ad stadard deviatio for a sample mea are: μx = μx x x = Also remember that if is ormally distributed the will be too, ad if is large, the sample mea will be approximately ormally distributed eve if is ot ormally distributed (Cetral Limit Theorem), that is ~ Normal( μ, ) approximately. What if your margi of error is too large? Here are ways to reduce it: Icrease the sample size (bigger ) [Populatio size does ot matter, as log as it is much larger tha the sample size.] Use a lower level of cofidece (smaller C) Reduce [Hard to do] x Choosig sample Size for the desired margi of error m: z * = (Solvig a simple equatio from m= z* ) m 9

Be careful: You ca oly use the formula ± z* uder certai circumstaces: Data must be a SRS (simple radom sample) from the populatio. Data must be collected correctly (o bias). The margi of error covers oly radom samplig errors. You must kow the populatio stadard deviatio. Examples:. A questioaire of drikig habits was give to a radom sample of fraterity members, ad each studet was asked to report the # of beers he had druk i the past moth. The sample of 30 studets resulted i a average of beers with stadard deviatio of 9 beers. a) Give a 90% cofidece iterval for the mea umber of beers druk by fraterity members i the past moth. * ± z = ± = ± = (9.3, 4.7) 30 b) Is it true that 90% of the fraterity members each moth drik the umber of beers that lie i the iterval you foud i part (a)? Explai your aswer. No. The C.I. is for, ot for the actual idividuals. c) What is the margi of error for the 90% cofidece iterval? * m= z = d) How may studets should you sample if you wat a margi of error of for a 90% cofidece iterval? z * = = m 0

Stat 5 Lecture Notes Chapter 8.5 ad. A sample of STAT 5 studets yields the followig Exam scores: 78 6 99 85 94 53 88 90 86 9 75 9 Assume that the populatio stadard deviatio is 0. The sample mea ca be calculated by calculator to be 8.83. a) Fid the 90% cofidece iterval for the mea score μ for STAT 5 studets. * 0 ± z = ± = ± = (78.08, 87.58) b) Fid the 95% cofidece iterval. * 0 ± z = ± = ± = (77.7, 88.49) c) Fid the 99% cofidece iterval. ± = ± = ± = (75.39, 90.7) d) How do the margis of error i (b), (c), ad (d) chage as the cofidece level icreases? Why? Margi of error will (icrease or decrease?) as cofidece level icreases. wat to be more cofidet that the populatio mea will fall i that rage eeds to be (bigger or small?). We rage. So the How does sigificace level α i a two-sided test relate to cofidece itervals? If you have a -sided test (a claim o if μ is equal a umber μ 0 or ot: H 0 : μ=μ 0 vs H : μ μ 0 ), - α = the cofidece level. For example, a 95% cofidece level would give you α = 0.05. If you have a -sided test, if the α ad cofidece level add to 00%, you ca reject H 0 if μ 0 (the umber you were checkig) is ot i the cofidece iterval. Graph below shows how to use a -α cofidece iterval to draw coclusio for a -side test.