Question

Part B: Normal Distribution Probabilities. For all questions, include the probability statement, a sketch, the calculator syntax, and the solution rounded to four decimal places. Ex.) Find the probability that a z-score is less than 2.02. 9.) Assume that annual pear consumption is Ples 2.02) normally distributed with a mean of 8.6 lbs. and standard deviation of 2.2 lbs. Find the probability -0.9783 that annual pear consumption is less than 11 lbs. 2.02 normalcd6-10000,2.02,0,1) 5.) Find the probability that a z-score is between -0.29 and 1.27 10.) Assume that annual pear consumption is normally distributed with a mean of 8.6 lbs. and standard deviation of 2.2 lbs. Find the probability that annual pear consumption is between 6 lbs. and 10 lbs. 6.) Find the probability that a z-score is greater than 1.87 7.) Find the probability that a z-score is less than -2.54 11.) Assume that annual pear consumption is normally distributed with a mean of 8.6 lbs. and standard deviation of 2.2 lbs. Find the probability that annual pear consumption is greater than 7.2 lbs. 8.) Find the probability that a z-score is less than -1.57 or greater than 0.23.
do all of part b please :)

Answer 1

According to the rules only first four subparts to be answered.

PC Between -0.29 and 1.21) = P(-0.292221.21) - P(241.27)-P (2<-0.29) = P(Z <1.27)-1+ P(Z <0.29) 1.27 = 0.8980-1+ 0.6141 -0.29 0 = 0.5121 (6.) Plz>187) = 1-P (22187) =1-0.9693 = 0.0307 عاد 1.81 7.) P(Z <=2.54)= HP(2<2.54) = -0.9945 = 0.0055 -2.54 l=0 8.) P(7.57 or >0.23)= P(23-1.57 or 220-23) = P(2<-1.572+ P(270-23) =l-PC2<1-57)+ IP(Z <0.23) - 4-0.9418- 0.5910 = 0.4672. -157 U-0.28