University:
Ridge Point High School
Course:
Algebra II
Academic year:

2020
Views:

481

Pages:

6
Author:

D. Marsden

NORMAL µ 02 book 2 chapter 3 DISTRIBUTION mean standard deviation variance symmetrical do't mÉa mean I median 681 of data lies between one standard deviation of 2 standard deviations of mean 951 i 3 standard deviations of Mean 99.71 Example I a mean D x 8 0.5 II variance b P 7.8 X 8.2 uomo each side 0.68 68 1 one 107 93 14 standard deviation 2 7 02 72 49 97.5 less than 70 the 2.51 greater than 70 T standarddeviation 5 100 95 5 2 12 standarddeviations 70 1072 5 70 10 10 10 60 FINDING PROBABILITIES normal cumulativedistribution normalCD example X N 30,42 9 PIX233 lower mean Mx 50 30 20 10 upper 33 33 3033 into calc X 0.77 b P x 224 PX 24 0.9332 C P 3351 138.2 0.1706 d P YC27 1 p 272 I YET tummiet or x 32 232 1 0.46 0.535 Ex 3B Xu N 30,22 1 a 3 1144333 Y N 25,25500 a PLY220 y o gq 5 b PIX 26 C P X 231 6 P x 31.6 0.2118 u 0.97 5 b P 1824426 0.499 C Ply 23.8 0.595 M N 15 1.52 t P Mc 14 I prev Ans 0.2520 0.7475 b sum of answers I as all probabilities add upto a i P M 14 y N 6.0.82 a i Pl X 7 it P XL 5 O 1056 V 0.1056 b PIX 5 0.1056 Ans 1.179 10 3 8 X a P XC72.87 0.0221 lower O upper 12.8 b PC12.92 2131 B 4010.68 P X 25 X gg I P XE25 INVERSE NORMAL to find a in p yea No area DISTRIBUTION on calc 0 probability one is perfect 0.722 9 9 22.023 b Pix a 0.4 Exca Pac 6 pcxggzg.gg 0.0921 PIX La 0.3 0.3921 0.3 9 19.178 a i P Y a 0.975 Ii PCYCb O 6 80.78 YCa 0.025 9 b 70.6 1 67 80.787 0.075 STANDARD NORMAL DISTRIBUTION mean is o IX N standard deviation I MOZ E You can STANDERDIZE 2 fancot 41 IN will be distributed mean 0 so 1 Plz La a fancccysymbown a 2 2 XI IL s 2 555507 53 451 is Finding N X N µ 32 PEX 207 0.2 find N need to use STANDARD L PLZ 20317 0.2 A inverse MEE 0.8416 2 2 0.8416 NORMAL 0 p Finding o X N 50,02 DC I'M o 507 0.2119 462 0.7998 inversetofind 2 find a PLAC467 0.2119 O gag 0 PLZ 4q 49558 2203 5 3 0.25 0 agg 18.6459 674 2051 3.37 200 4 203.37 N 203mF Binomialto Normal NBLhip if his large N Np G and pisclosetoof Fptp you can convert to normal HypothesisTesting look at MEAN of a sample from population X N N 02 T Sammy N NIE on in Ho N 60 assume Ho X N 60,32 NL 60 H normal CD Faith I no like usual f 1gj If you need to find NINE Ho N 0.580 Hi N 0.380 o.it 0.1151 70.05 sothere is notenough evidence to rejectHo and conclude the mean amount ofjuice isless than 60m relate it backtoquestion a you can standardize I pctcsa.n CRITICAL value 2 standardize YEI D N 0.58 0.0152 D region you test statistic N 0.5810 tailed s n'ticalregions ends we have to find a critical region so we standardise at two inverse gg normal an 2.5758 decode data 2.5758 2.5758 II 0.58 0.574 N 100,152 X a AIGHT I Ho N 100 Hi p 100 X N 100,152 2 Eff Iggy to 1959 1 asa Iggy critical region I b 10252103.29 notsufficentevidence to rejectHo cannotsaychocolate improves IQ 5 1.959 7807 100 103.285 103.29