Question

The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds produced from the plant was analyzed for a particular collagen. The collagen amount was found to be normally distributed with a mean of 61 and standard deviation of 9.8 grams per mililiter. a. What is the probability that the amount of collagen is greater than 60 grams per b. What is the probability that the amount of collagen is less than 90 grams per c. What percentage of compounds formed from the extract of this plant fall within 1 mililiter? mililiter? standard deviation of the mean?

Answer 1

Solutions - The normal distribution with Given - M = 61 = 9.8 @ Find P6 x 760) | P(x760) = 9 - P(x <60) - 1-P (Z L 60-61) = 1 P(Z L - 0.10) .... Use a table = L - 0.4602 ZP go-611 9.8 P(x > G 0) : 0.5398 6 find P(x490) | PCX L 90) = P ( x-u 90-1) - PC2% 9001) = P(Z. 42.96) ... use a table P(x<902 = 0.9985 © Eind. P within 1 standard deviation of the mean) = P(u-6<x < uto = P(61-9. 8LX C61+9.8. - P(51.2 < X < 70.8) P(x < 70.8) - P(x (512)

(512 للx) ۴ - ( 108 ) = ای - 2 51 . 2 ) 2 - ( 70 . 861 ح)P المهلت ( 2 ) - (2 1 . 00 ) م table 2ل.---- 0 . 47 - 0 . 343 0.6826 TPC within or the mean) 1 standard deviation 6826.م که