Question

AL CURVE WORKSHEET HW Peor BRUNT OIVEN 4 NacnAL 7STRIBUTION OF EMLE . l-190 and 1= 15 FOR AMERS (a) FIND THE PERCENT OF LEVELS ABOVE 193 RIBUTION OF CMLESTEROL LEVELS MITH Enl = 15 (For AMERICANS). LRaw 70 NEMET HORRETH (6) BELOW WHAT LEVEL ARE 12% OF THE PEOPLE? (c) WHAT'S THE Z SCORE OF THE CHOLESTEROL LEVEL WHICH HAS 60% OF THE LEVELS BELOW IT? FIND THE CHOLLSEROL LEVEL THAT HAS A PERCENTIVE OF YO. ANK e) IF ONE PERSON IJ SEZECT AT RANDOM, WHAT'S THE PROBABILITY HIT (HER CHOLESTEROL LEVEL IS BELOW 164?

Answer 1

Solution? Given that, u=190 6 =15 Plx>193) =1-PC x< 193 -P - - <193-190 15 - - go. z1-p [z c 0.2] - 0.5793 = 0..4207 P(x>193)=0.4207 probability =0.4207

Date. 2) do is The zo distribution of the 12 Plz 2 z )= 12 olo Plz < z ) = 0.12 P(Z <-1-18) = - 1.18 z = -1:18 using 2-score formula X=2G tee x= -1.18 X1 5+190 x- 172 .3 x-17253 c) the 2- distribution of the 60% is PL z cz) = 60 0% Pl2 <z)= 0-60 PCZ50.25) = 0.60 2 -0.25 using 2-score formula x=26tu x=0.25x15 X- 193. 75 +190 z score =0.25

P(x470) =P [** < 50-140] =P I z<-8 = 0.00 -0 % P(x270) = 0% P( x4164) = P X -4 < 164 - 190 e) is =P I z <-1-73] = 0.0418 | P ( 3< [64) = 0 - 048 probability 2 0.0418