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 79 and standard deviation of 6.8 grams per mililiter.

(a) What is the probability that the amount of collagen is greater than 77 grams per mililiter? answer:

(b) What is the probability that the amount of collagen is less than 73 grams per mililiter? answer:

(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean? answer: %

NormalRandomVariables

Answer 1

Solution :

Given that,

mean = = 79

standard deviation = =6.8

P(x > 77) = 1 - P(x<77 )

= 1 - P[(x -) / < (77-79) /6.8 ]

= 1 - P(z < -0.29)

Using z table

= 1 -  0.3859

probability= 0.6141

(B)P(X<73 ) = P[(X- ) / < (73-79) /6.8 ]

= P(z <-0.88 )

Using z table

= 0.1894

probability=0.1894

(C) P( - 1< X < + 1) = 68%

answer=68%