Question

I have the wrong answers and cannot figure it out

f pairt tente k twee-nonih penind was t Suppcse we rando nly select a sampe of 100 blue collar workers. Dased on Maricccthios estimates: (a) What is the protabilty that the average amount of paid me los: during a three month period fer the 100 blue collar workere wil exceed 1.5 days? (Round means to 1 decimal place, standard deviations to 2 decimal places, and probabilities to 4 decimal places. 14 b w i l is the p obabilit t al he average amaunl of uripaid ime l during a hi e month perico er the 100 blue colla work s wil e ceed 1 Ca s? Round standard dev at ons to 2 dec mal places and Probabil ties to 4 decimal places 05 pMeviausly eslimaled 14 days? (Click lu eeleci) Ί , the probablity of observing the sample 15 (Click lu select r the mean is acully 1.4 Reterences Book Resources

Answer 1

$Let~X~be~a~random~variable~denoting~amount~of~paid~time~lost~during\\ ~three~month~period.\\ Here,~assume~X\sim N(1.2,~1.4)\\ Then~\bar{X}\sim N(1.2,(1.4/10)^2)~or,~\bar{X}\sim N(1.2,(0.14)^2).\\ \mu_x=\mu=1.2\\ \sigma_{\bar{x}}=0.14\\ P(\bar{X}>1.5)=1-\Phi\left(\frac{1.5-1.2}{0.14} \right )=0.0161\\ (b)~\mu_x=\mu=1.4\\ \sigma_{\bar{x}}=1.3/10=0.13\\ P(\bar{X}>1.5)=1-\Phi\left(\frac{1.5-1.4}{0.13} \right )=0.2209$

$No,~the~probability~of~observing~sample~is~0.2209.$