Chapter 6: Normal Probability Distributions

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Chapter 6: Normal Probability DistributionsSection 6.1: The Standard Normal DistributionContinuous Probability DistributionsDefA density curve is the graph of a continuous probability distribution.Requirementsi.e. P ( x) 11.The total area under the curve must equal 1.2.Every point on the curve must have a vertical height that is 0 or greater.Uniform Probability DistributionEx: The bus to Union Station leaves every 30 minutes and is uniformlydistributed. Find the probability that a randomly chosen person arriving ata random time will wait between 10 and 15 minutes?𝑃𝑃(10 𝑥𝑥 15) Normal DistributionsDefA continuous random variable has a normal distribution if its density curve is symmetric and bell-shaped.Specifically, the curve is given by: y 1σ 2πe 12σ 2( x µ )2(Don’t worry, we’ll never use it.)Ex: The weights of all firefighters are normally distributed with a mean of 200 lbs and a standard deviation of 7 lbs. What’s theprobability that a randomly chosen firefighter weighs between 185 and 195 lbs?:Standard Normal DistributionDefThe standard normal distribution is a normal probability distribution with µ 0 andσ 1.

Important Notes1.The z-score is used on the horizontal axis.2.The area of the region under the curve is equal to the associated probability of occurrence.Two Ways to Find Area1. Use Table A-2.Look up the area under the curvethat lies to the left of z-score (may firstneed to convert data to z-score).2. Use Graphing Calculator (TI-84 Plus)(a) 2nd VARS DISTR(b) normalcdf ( lower , upper , µ , σTwo Ways to Find Z-score1. Use Table A-2.Look up the z-score associatedwith the area that lies to left.2. Use Graphing Calculator (TI-84 Plus)(a) 2nd VARS DISTR(b) invNorm( area , µ , σ , Tail )Ex:Find the probability given the following z-scores for a standard normal distribution.(a)P ( z 1.35)(b)P ( z 0.68)(c)P ( 2.43 z 0.88))

Ex:Find the z-score associated with the 15th percentile.Ex:Find the probability given the following z-scores for a standard normal distribution.(a)P ( z 0.55)(b) Would P ( z 0.55) differ from (a)?Ex: Find the z-scores that separate the top 10% and bottom 10% of all values.*Specific Notation: 𝑧𝑧𝛼𝛼 is the critical value that denotes a z-score with an area of 𝛼𝛼 to its .Ex: Find 𝑧𝑧0.05

D-Section 6.2: Real Applications of Normal DistributionsZ ScoresI IA z score is the number of standard deviations that a given value x is above or below the mean.Def①Sample: z Formula:x xsx µσ§ 6i ;one." Round-Off Rule: Round z scores to two decimal places.[ii.z Population:Ex: Consider your height in inches. Calculate the standardized value (z-score) for your height given that in the UnitedStates the average height for women is 63.7 inches with a standard deviation of 2.7 inches and for men is 69.1 inches witha standard deviation of 2.9 inches. Would you be considered tall for your ��� X-height o f m e n i . u sDa x x varies by person(Jorge)6 8 inches6821691 1-0.382Z"'t" i'%% ; §.- 0 9E FNDWhat is the probability that someone of your gender- is taller than you?cdf(68,6%1, 6941,259) normal 0.5-Ex: The average for the statistics exam was 75 and the standard deviation was 8. Andrey was told by the instructor that hescored 1.5 standard deviations below the mean, and the scores were normally distributed.students scored higher?XtiAdyFwhatpe@r.fUtexI t z s75 H e belowt f-1.5 ) (8)163-7 0.933093-3%0(603,71 55,08)norwalcdf 5,I ( x 637 0.5 M(note:negative.''#z -1.5),',,630G-Ex: The life spans of a brand of automobile tires are normally distributed with a mean life span of 35,000 miles and a- ofo a randomly selected tire is 34,000 miles. Find the z-score of this tire.standard deviation of 2250 miles. The life spann a randomly selected automobile tire has a life-span less than or equal to 34,000 miles?Can you find the probability that-ta-②lifespanof t i r e s i n w i l e s . X 3 4 ,oo on iZ-score:34000) io .-iz 2-[(xE oxjx #(34ooo-35000)225 - D(34k X c35K)57 I I,44k(34900,35%00,35%00,22550) 04,3285o .5-normalCdf

lo-z-score)When to use Normalcdf? -f i n d proddataTo find a probability when the ( µ , σ , x ).are given.a -F i n d X-valuesWhen to use- InvNorm? AbWhen the percent or area or probability is given and weexpandingvalues.are trying to find the .*OEx: The completion times to run a road race are normally distributed with a mean of 190 minutes and a standarddeviation of 21 minutes.-OE ( x 150) 1214490) - l ? (1502 2190)(a) What is the probability that a randomly selected runner will finish the race in-less than 150 minutes?0.5-normal cdf(150,190,190,21)1 - ,iii-]1 -150(b) What percentage of runners will finish the race between 205 and 245 minutes?17/2052 2245) normaledf(205,245,190,21) :&190"k 4.Ex: A construction zone on a highway has a posted speed limit of 40 miles per hour. The speeds of vehicles passingthrough this construction zone are normally distributed with a mean of 46 mph and a standard deviation of 4 mph. (a) What percentage of vehicles exceed the speed limit?R46)tR(402 246)R(x 4)/x normolcdf(40,46, 46,4) o-s t 0.933 I38§142 4 6s o40540(b) If the police wish -to ticket only those drivers whose- speed falls in the upper 20th percentile, what is the minimum speedof a driver that will be ticketed?Given g ) X - i n v Norm(speed)0-2, find X-valvepnrtail( 0 . 2 , 46,4, RIGHT)-49 1%9%1%4. speed'cut-off

nuts! huge!/-fo r daterets,Section 6.3: Sampling Distributions and Estimators- o "EI{nuttieoretical (typically represented as a probability distribution in the format of a table, histogram, or formula)Sampling DistributionDef The sampling distribution of a statistic is the distribution of all values of the statistic when all possible samples ofthe same size n are taken from the same population.Ex: Given three pool balls we will select two of the balls (with replacement) and find the average of their numbers.a l l samples of 2 pool ballsI t mean#of 2 selectedballs.{ 1 1, 12,13,.} (a) Fill in the table to find 𝑋𝑋 the average of a sample of size two. (b) Fill in the table below using the data from (a).sampleOutcomeBall 1Ball 2MeanDistribution MeanFrequencyRelative'' I1.1Frequency2.1liO n i3.1NEWi4.20.22224105 2.205 5.223sampling0.33J6.2I27.30.22TI23-5#Distribution 38.3'o - n i9.3}&2/}9}85 ②(c) Draw the relative frequency distribution.look, appox normal-symmetrics t a s h : mean IYa1.5-3/92.511gµ y, \ i #0.30.2Mi0.1-(sample)9i 15 I 1.sbAs the number of samples approaches infinity, the relative frequency distribution will approach the sampling distribution.f\Sampling Distribution- of the Mean1-theoreticalE x above: pop N 3, Sanden-2The sampling distribution of the mean is the distribution of sample means, with all samples having the samesize n taken from the same Important Notes1.The sample means target the value of the population mean(i.e. the mean of the sample means is equal to the population mean)2.The distribution of the sample means tends to be a normal distribution.(Normal)(The distribution tends to become closer to a normal distribution as the sample size increases.)An unbiased estimator is a- statistic that targets the value of the corresponding population parameter in the sensethat the sampling distribution of the statistic has a mean that is equal to the corresponding population parameter.DefIe'd?'raje,?:{" ' " f - "Phat"Unbiased estimators:Biased estimators:

Section 6.4: The Central Limit Theorem*Central Limit TheoremM M- mean µ and standardI deviation σ .o 1. The random variable x has a distribution withGiven(the distribution may or may not be normal)0 2. Simple random samples of size n are selected from the population.Conclusionsdist.normalThe distribution of sample means x will approach aas the sample size increases.0 1.- CO2.m q i !The mean of all sample means is the populationI meanM.µµ T #Mex3. (always)The standard deviation of all sample means is given byf µ *a . " s t .devotdist"σx Important'changes a snone o fAnimation: sim/sampling dist/ (simulation app)samplingNotes About Distributions onchanges!any1. If the original distribution is normally distributed, then for any sample size n , the sample means will be normallydistributed.2. If the original population is not normally distributed then for samples of size n , the distributionof sample means can be approximated by a normal distribution.C OEx: Find 𝜇𝜇𝑥𝑥̅ and 𝜎𝜎𝑥𝑥̅ for the given distributions.sampling'} oo#normal(a) Given a normal distribution where 𝜇𝜇 10, 𝜎𝜎 3 and 𝑛𝑛 9.µ f&.-Sampling m a nf %careful!samplingi hNorths t . Dev.3/3 1r #Sapling3/59ti, %normallydist(b) Given a distribution with 𝜇𝜇 77, 𝜎𝜎 14 and 𝑛𝑛 49. 3o #Lot49 M y M 7 7 § µ, .Of o f f - 14µg '4/7 2 7 STEPS to Use Sampling Distribution for 𝒙𝒙requirements)(check1 : Check if 𝑥𝑥̅ is Normal (or is 𝑛𝑛 30? )(compute)2nd: Find 𝜇𝜇𝑥𝑥̅ and 𝜎𝜎𝑥𝑥̅rd3 : Use normalcdf for the rest just don’t forget to use information from step 2st(justlike6-1,6-2)

c a n u s e-a n y§samplesize!nEx: The GPAs of all students enrolled at a large university have a normal distribution with a mean of 3.02 and a standarddeviation of 0.29.1,62pwayoriginal-§6.(a) Find the probability that one randomly selected student will have a GPA greater than 3.20. [(x 3.20) 11 73.02)-12/3.022 23.2)baN',5'0.5-normal cdf(3-02,3-7,102,0-29) i' 7'.I,3%313%2 3.2f(b) Find the probability that 25 randomly selected students will have a mean GPA greater than 3.20.I n - 2 5 Note Housing\Samplingaousesampling1'/ 0j29 o-058 ,µ- µ 3.02is:*:*:anNI % D-( I 3 .2) i [ ( I 3.02)- I (3.022 23.2) 95603 .o z(c) Find the probability that 10 randomly selected students will have a mean GPA between 2.90 and 3.10.&*sampling!n - l o useanswer[(2.9027 3.10)/ cdf(299,31,3 12o.FI/fo) o#normal*/,Ex: Let 𝑥𝑥 𝐶𝐶𝐶𝐶𝐶𝐶 salaries (in thousands) where 𝑥𝑥 is skewed right with 𝜇𝜇 139, 𝜎𝜎 45.n o"A i(a) If all possible random samples of 40 CEO salaries (in thousands) are taken, how would you describe the- standard deviation of the' sample distribution be?distribution of sample means? What would thea."Y:&.("bycentrollimittheoremn-40530?Q -Te own n o t normoldist- need n 3 045/540 7.115.afflthousandddt

(b) What is the probability a sample of 40 CEOs make between 137 and 145 thousand dollars?OH{ME I fdist samplingsantirs: Moe.IS?. *iro.cdffitt,i normal5)ELMcIlialust, *\{ I , FF,f o )0 -(c) Given a sample of 40 CEO’s salaries, at what salary does the top 10% of CEO salaries begin at?I I u s e sampling( p , µ , r , tail)i n v Nor m(o-t, Ia,"451%,RIGHTinvN o r m1 4 8.1 1 8. . . .oolotop'O" I"In.A* ?148.1"with a sample of 4 0 L E O salaries, t h e top 1 0 % o f t h e i r salaries( m e ) beginsa t 148.1 thousand dollars."(d) What is the probability a sample of 40 CEO’s makes less than 120 thousand dollars? Is this unusual?

Chapter 6: Normal Probability Distributions Section 6.1: The Standard Normal Distribution Continuous Probability Distributions Def A density curve is the graph of a continuous probability distribution. Requirements 1. 1The total area under the curve must equal 1. i.e. Px 2. Every point

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