Question

9.8) In problem 9.7, to what value must we reduce the standard deviation of jelly bean weight so as to make the probability of interest equal to .95?

Answer 1

Answers CD n=25 * Given collect from the data * let us assume that the data as followed * The jelly beans have a mean weight of 5 grams with the S.D of 1 gram & to find out the pr (X>, 407) where Now M=5 and 6=1 - 0 * we have to sampling distribution of sample mean it would be nomal distrubution is Mean = llx=l=5 # SoD sample mean is 5= = 0.2 Now at have to find out z -score za ve yen = 470 - = -150 vo then fr (T>, 4.7) = Px (27, 1650) and Pr (2>, 4-7) = 1- Pr (221:50) then z = -1.5 and 0.00 and its area can 0.0668 NOW Prl=>, 4.7) = 1-pr(24-1.50) > 1-0.0668 → 0.933 2 (9.8) * it is the values of the bean weight has can reduce make the the standard deviation probability of interest equal to 0.95

Now u=5gm, o=1gm, n=25 then plu>, 4.7) = 0,95 = P(X24.7) = 0.05 Pleyt Mac or J20.05 > PL zz 40706] -0.65 → P[z_ts] = 0.05 then -15 -1.6648 = 5= 0.9119 0.0881 in the probability oby will reduce then interest has F then 0.95 THE END - >