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# STAT 3360- Normal Distribution STAT 3660

WMU

GPA 3.29

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## Popular in Intro to Statistics

## Popular in Statistics

This 6 page Class Notes was uploaded by Laura Romero on Sunday October 11, 2015. The Class Notes belongs to STAT 3660 at Western Michigan University taught by Andrews in Fall 2015. Since its upload, it has received 70 views. For similar materials see Intro to Statistics in Statistics at Western Michigan University.

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Date Created: 10/11/15

OOO STAT 3660 Normal Distribution To visualize the data when nding the probability of data Draw a histogram Draw a smooth curve on it Erase the bars Smooth curves are used to represent the distribution of the population of data Quantities can be studied compared and discussed by being represented by curves Curves have many different shapes The most wellknown curves are the bellshaped curves which has a de nite shape given by a formula Bellshaped curves are considered normal distribution Most of the data values lie in the middle of the range of data The mean the center of the curve and the standard deviation the spread of the curve are the two key features of the bell curve To determine the area under a portion of the curve one must change the value wanted to the Zscale standard normal scale The value x is converted to the Z scale by the equation 2 x mecmSD which nds the area of Z The Z table gives the area next to the value of the z score Elma HSHH HSHr r HSHE HS 2 HS i5 HS2H HS24 HS2E HS52 HSSlS HS4H HS44 HS4E HSS2 HSSilS HSilSH HS54 HS5 HS 3quot HS TS HS TE HS HS ET HSE HSES HSEE H5HS H5Hi5 HAS il HAS 4 HE E H522 H525 H525 H553 H514 H544 H54E H5S2 H5SS HESE HAS TH HE T4 HAS Tquot HAS E HE E4 HAS EE H551 HEES lms H E42 ill quotF4S ill E453 ill ESE ill quotFSS ill ESE ill 55 H H 35 ill 3 H il E TS H 3quot T5 H E TE ill E H quotF ill 3quot EE ill TE ill ill 55 H EHH ill EH2 H EHS H EHE HE HE HE 15 HE E ill E2 H E24 H E25 ill H ES H H il H E4 H il ill E4E H ES H HESS ill ESE H Eilan H il ill E5quot ill HE T HE TS H E TS H E Tquot 5T 55 Elma HEHT HEHE HE 1H HE 1 HE IS HE S HE 15 HE E HE E HEE HEEE HE24 HEES HEg2i5 HE2E HEg ZE HES HES4 HESquot HESE HE4 HE42 HE4S HE44 HE4S HE45 HE4T HE4E HE4E HES HESE HESS HES4 HES4 HESS HESlS HES HESE HESE HEilSH HEE HE52 HEi52 HEiliS HE54 HEELS HE55 HEi55 El 1 55 255 251 252 255 254 255 255 25 2 255 215 21 1 212 21 5 214 21 215 21 215 215 225 221 222 225 224 225 22quot 225 225 255 251 252 255 254 255 25quot 255 255 245 241 242 245 244 245 245 24 2 245 255 lms HE T HE T HE TE HE quotFE HETE HETE HE EH HE EH HEE HEE HE HE E2 HE HE HE ES HEE4 HEE4 HE HE HE HEEilS HEElS HEEilS HE E HE Equot HE E HE EE HE EE HE EE HE EE HE EE HE EE HEEH HEEH HEEill HEEH HEE HEE HEE HEE H552 H552 HEEE HEEE H552 HEES HEES HEES HEES HEES HEE4 HEE4 E 255 25 255 255 2755 251 252 25 254 255 255 25quotquot 255 255 255 251 252 255 2 255 25 2 255 255 251 252 255 2 255 25quot 2 255 555 551 552 555 555 555 55 555 515 511 512 515 514 515 515 51 Eran HEEil39 HEEi HEEil39l HEEil39E HEET HEET HEET HEET HEEquot HEET HEET HEET HEEquot HEET HEET HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE OOOOOOOOO IDJEIIEIT 251 1994 31 IDEDD IDDIEIEEI 252 1994 313quot DEGquot DEED 253 1994 323quot DDDD IDJEIDD 71994 321 71999 IDJEIT ID 1995 322 IDEDD DJ 253 1995 323 DEGquot DDT 25 1995 17quot ID 2 71995 325 17quot 71933 1995 323 IDEDD 1933 233quot 1995 32 DEGquot 71934 2431 71995 3 71999 71934 2432 IDquot 39DD 323quot 71999 33 71935 2433 71931321 333quot IDID 3quot IDDTID 234 DDDD ID 235 DJEIDD quotDIE2 DDDD quotDIE3 IDEIDE quotDIE4 IDJIDE IDJIDD quotDIEquotDi DJ E quotDIET IDJ39 D quotDquot D 539 l D quotDIE 3quot DEED quotDIEDquot IDJ39 331 IDJ39 ED quotD32 IDESE 333 IDJ39i D quotD434 DEEquot3quot 1 D 313quot 1 D iii 1 393quot D 211quot D 21 D 3 22 ID itquot 2 3 D ED 24 D 2 71394 23 D 2 D 2 D DIDD 23quot D DID 33quot D DIDD 31 D DIDEI h wmm Eaieiaiahi mimI mmu I V 5n JdJJJdJJJdJJde All other types of area can be found by using the symmetry the 2 curve To nd probabilities Solve for Z using 2 x meanSD x is the number given Once you have Z nd the Z on the Z tabe Look for its area which is the probability in decimal form To nd percentiles Look for the percentage in decimal form under area on the Z tabe Enter the 2 value in the equation 2 x meanSD OR X mean Z SD Solve for X STAT 3660 Normal Distribution 0 To visualize the data when nding the probability of data 0 Draw a histogram 0 Draw a smooth curve on it Q Erase the bars 0 Smooth curves are used to represent the distribution of the population of data 0 Quantities can be studied compared and discussed by being represented by curves 0 Curves have many different shapes 0 The most wellknown curves are the bellshaped curves which has a de nite shape given by a formula 0 Bellshaped curves are considered normal distribution 0 Most of the data values lie in the middle of the range of data 0 The mean the center of the curve and the standard deviation the spread of the curve are the two key features of the bell curve 0 To determine the area under a portion of the curve one must change the value wanted to the Zscale standard normal scale 0 The value x is converted to the Z scale by the equation 2 x meanSD which nds the area of Z 0 The Z table gives the area next to the value of the z score Elma HSHH HSHr r HSHE HS 2 HS i5 HS2H HS24 HS2E HS52 HSSlS HS4H HS44 HS4E HSS2 HSSilS HSilSH HS54 HS5 HS 3quot HS TS HS TE HS HS ET HSE HSES HSEE H5HS H5Hi5 HAS il HAS 4 HE E H522 H525 H525 H553 H514 H544 H54E H5S2 H5SS HESE HAS TH HE T4 HAS Tquot HAS E HE E4 HAS EE H551 HEES lms H E42 ill quotF4S ill E453 ill ESE ill quotFSS ill ESE ill 55 H H 35 ill 3 H il E TS H 3quot T5 H E TE ill E H quotF ill 3quot EE ill TE ill ill 55 H EHH ill EH2 H EHS H EHE HE HE HE 15 HE E ill E2 H E24 H E25 ill H ES H H il H E4 H il ill E4E H ES H HESS ill ESE H Eilan H il ill E5quot ill HE T HE TS H E TS H E Tquot 5T 55 Elma HEHT HEHE HE 1H HE 1 HE IS HE S HE 15 HE E HE E HEE HEEE HE24 HEES HEg2i5 HE2E HEg ZE HES HES4 HESquot HESE HE4 HE42 HE4S HE44 HE4S HE45 HE4T HE4E HE4E HES HESE HESS HES4 HES4 HESS HESlS HES HESE HESE HEilSH HEE HE52 HEi52 HEiliS HE54 HEELS HE55 HEi55 El 1 55 255 251 252 255 254 255 255 25 2 255 215 21 1 212 21 5 214 21 215 21 215 215 225 221 222 225 224 225 22quot 225 225 255 251 252 255 254 255 25quot 255 255 245 241 242 245 244 245 245 24 2 245 255 lms HE T HE T HE TE HE quotFE HETE HETE HE EH HE EH HEE HEE HE HE E2 HE HE HE ES HEE4 HEE4 HE HE HE HEEilS HEElS HEEilS HE E HE Equot HE E HE EE HE EE HE EE HE EE HE EE HE EE HEEH HEEH HEEill HEEH HEE HEE HEE HEE H552 H552 HEEE HEEE H552 HEES HEES HEES HEES HEES HEE4 HEE4 E 255 25 255 255 2755 251 252 25 254 255 255 25quotquot 255 255 255 251 252 255 2 255 25 2 255 255 251 252 255 2 255 25quot 2 255 555 551 552 555 555 555 55 555 515 511 512 515 514 515 515 51 Eran HEEil39 HEEi HEEil39l HEEil39E HEET HEET HEET HEET HEEquot HEET HEET HEET HEEquot HEET HEET HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE HEEE IDJEIIEIT 251 1994 31 IDEDD IDDIEIEEI 252 1994 313quot DEGquot DEED 253 1994 323quot DDDD IDJEIDD 71994 321 71999 IDJEIT ID 1995 322 IDEDD DJ 253 1995 323 DEGquot DDT 25 1995 17quot ID 2 71995 325 17quot 71933 1995 323 IDEDD 1933 233quot 1995 32 DEGquot 71934 2431 71995 3 71999 71934 2432 IDquot 39DD 323quot 71999 33 71935 2433 71931321 333quot IDID 3quot IDDTID 234 DDDD ID 235 DJEIDD quotDIE2 DDDD quotDIE3 IDEIDE quotDIE4 IDJIDE IDJIDD quotDIEquotDi DJ E quotDIET IDJ39 D quotDquot D 539 l D quotDIE 3quot DEED quotDIEDquot IDJ39 331 IDJ39 ED quotD32 IDESE 333 IDJ39i D quotD434 DEEquot3quot 1 D 313quot 1 D iii 1 393quot D 211quot D 21 D 3 22 ID itquot 2 3 D ED 24 D 2 71394 23 D 2 D 2 D DIDD 23quot D DID 33quot D DIDD 31 D DIDEI h wmm Eaieiaiahi mimI mmu I V 5n JdJJJdJJJdJJde 0 All other types of area can be found by using the symmetry the 2 curve 0 To nd probabilities O Solve for Z using 2 x meanSD x is the number given 0 Once you have Z nd the Z on the Z table 0 Look for its area subtract from 1 this will be the probability in decimal form 0 To nd percentiles 0 Look for the percentage in decimal form under area on the Z table 0 Enter the 2 value in the equation 2 x meanSD OR X mean Z SD 0 Solve for X

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