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Elementary Statistics

by: Hilbert Denesik

Elementary Statistics STAT 200

Hilbert Denesik
Penn State
GPA 3.92


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This 0 page Class Notes was uploaded by Hilbert Denesik on Sunday November 1, 2015. The Class Notes belongs to STAT 200 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/233132/stat-200-pennsylvania-state-university in Statistics at Pennsylvania State University.


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Date Created: 11/01/15
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curve The xmnbmirmrmd was a nurmal curve made furzrscures 1h lhls meuel themanls n and thes39andard demahunlsl abuut helghls le huurs ersleep many utherlmnahle Flndlng Areas Under the Standard Normal curve 0 The lame m1ng 232 unhepaskel shuws lhe mam Lhzleflnfz meme 0 Spreadsheelpmgams hke Excel have eemmahuslhal aleulalehemqal curve eras nemakumznnmmalzmmGalv plwnnn l w 40 PM Eagles l o Gmphmg calculaturs llke me T1783alsu have cummands that an beused Solvlng Normal curve Problems The Basics39 sechun o Themmmsu7huurs o The slanderd demahun l5 7 17 huurs Howto Flnd the Propomon Less Than a Value Tu ndme pmpumun ufa pupulauun wllh a lame less man a spen edmlue l Celeulelee zrscure furlhe spe ed mlue leftufthezrscure Exam Abeulwhalpmpemen ufsludents sleepless man s huurs em weekmgm ms wlll be me araluthele ufshuurs Plupumun Sleeplng Less than 5 Huuls Ale la le m5 e pmpmnleslms 1234537891D111213 Huuls elsleep Thamlmjnn 1 Calculate me slanderdlzed scare fur s huurs ufsleep The funnulals vlemuvRHumznnmmalzmmom pmmm l w 40 PM 835mg 1 value 7 mean z FurShuurszSr7177217 118 11152111213212 un page 2321 ndthearaluthele ufrl 18 Themswms 21mm 11 12 21mm 12 Abuut 12 ufcullege studems sleep less man s huurs Bu 2 week mgm Shuurs1slr ll 88 Howto Find the Propomon GreaterThan a value Tu ndme pmpumun Ufa pupulauun wnh a wine Wta thana spen edwlue 1 cm a Home furthe speci ed mlue Determme me am 1 me 1211 mm Home 3 Araluthengqt amlu helem Examplz Abuulwhalpmpumun ufstudents sleep mure man 111 huurs ma week mgm Tms15 me am tutherighlufl huurs Pvupumun S122pmg Mme 1m 1m Huuvs Ave 0 1111M m m 15 mmquot mm mquot m 2345578 9111111213 Huuvs uvs1eep Th2 mllnjnn n1 mumumznnmmmm 5 m pmmm 1 w 40 PM 1mg 1 Calculate Lhezrscure fur 1n huurs Z1ne71731717 2 Fmdmeammmemnerzzna Theanswmsabemnga Usep 232 3 Calculale answa39m Step 2 1 719511 n4abeu14 AbuuH Vaufcullegesludemssleep mnmvhanl huurs ermgm Howto Find the Propomon m an Interval Tu ndme pmpum n Ufa pupulauun 11131 lls ma mended 1mm 1 Calculates zescure fur ach md erme 1mm 2 Fmd me am pmbablhly 1 me le er ach zescure 3 021mm me mffa39mce belwem me Wm pmbabllmes On aweekmght what p cm tags ersmdems s1eep belwem s and 111 mm Th151s me am between 5 huurs and 111 huurs Pvupumun Hemeen 5 and 1m Huuvs 2345578 9111111213 Huuvs evs1eep Them111nm We lmed mfmmauun 21mm 5 and 1m huurs urs1eep 1n the prevmus Wm su1uuuns 1 Furl huurszl7 FurShu L118 2 Tu the left ufz e 76 me ara1s 21mm 1 96 vieluukummnmmaummwmvMwnnn1w 40 PM Basics 1 To the left ofz 118 the area is about 012 3 The answer is 096 012 084 About 84 of college students sleep between 5 and 10 hours on a week night How to Find a Percentile The term quotpercentile rankquot refers to the area probability to the left of a value For instance we found that the area to the left of 5 hours of sleep is 01190 This means that o the percentile rank for 5 hours is 01190 about 12 o the 12th percentile is 5 hours To find the value corresponding to a specified percentile rank 1 Determine the zscore that has the given percentile rank 2 Figure out what value has that zscore Example What is the 98th percentile of the distribution of hours of sleep The probability to the left of the answer is 098 The solution 1 Find the zscore for which the area to the left is 098 Look for 098 under ProbltZ in the table on page 232 o The value under Z Score is 205 The 98th percentile is 205 standard deviations above the mean 2 Determine the sleep value that has a zscore of 205 fileDRHOstat200normal2htm 7 of 9 3142000 11949 PM Basics 1 o 205 standard deviations is 205 17 35 hours 0 35 hours above the mean of 7 hours 105 hours The 98th percentile of the sleep distribution is 105 hours A formula to calculate the value that has a known zscore is XZ0u For 2 205 X 205 17 7 35 7 105 hours The Empirical Rule and Outliers The following three statements constitute the empirical rule 1 About 68 of the values in a population described by a normal curve are within one standard deviation of the mean 2 About 95 of the values in a population described by a normal curve are within two standard deviations of the mean 3 About 997 of the values in a population described by a normal curve are within three standard deviations of the mean The empirical rule is also called the 68 95 99 7 rule Example The number of hours that college students sleep on a week night is approximated by a normal curve with a mean of 7 hours and a standard deviation of 17 hours 0 About 68 of college students sleep between 53 and 87 hours on a week night The calculation is 7 i 17 0 About 95 of college students sleep between 36 and 104 hours on a week night The calculation is 7 i 217 0 About 997 of college students sleep between 19 and 121 hours in a week night The calculation is 7 i 317 When is a Value an Outlier There is no universally accepted criterion for declaring a point to be an outlier but most data fileDRHOstat200normal2htm 8 of 9 3142000 11949 PM Basics 1 analysts are suspicious when a data values zscore is not between 3 and 3 The motivation for this guideline is that about 997 of the values in a population described by a normal curve are within three standard deviations of the mean In other words nearly all zscores are between 3 and 3 Example In the Penn State sleep survey one student s response was that he had slept 16 hours the previous day Let s see where 16 hours falls in the normal curve model for sleep zl67 l79l7529 This zscore is outside the range 3 to 3 so we call it an outlier We can also use the table on page 232 to find the probability to the left of z 529 The closest zscore in the table is z 5 The probability to the left of this zscore is 09999997 or in percent terms about 9999997 In other words 16 hours of sleep is around the 9999997th percentile of the distribution Quite an accomplishment fileDRHOstat200normal2htm 9 of 9 3142000 11949 PM


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