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 77 and a standard deviation of 5.5 grams per milliliter.

(a)  What is the probability that the amount of collagen is greater than 67 grams per milliliter?

answer:

(b)  What is the probability that the amount of collagen is less than 90 grams per milliliter?

answer:

(c)  What percentage of compounds formed from the extract of this plant fall within 3 standard deviations of the mean?

answer: %

Answer 1

Solution :

Given that ,

mean = = 77

standard deviation = = 5.5

a) P(x > 67) = 1 - p( x< 67)

=1- p P[(x - ) / < (67 - 77) / 5.5]

=1- P(z < -1.82)

Using z table,

= 1 - 0.0344

= 0.9656

b) P(x < 90)

= P[(x - ) / < (90 - 77) / 5.5]

= P(z < 2.36)

Using z table,

= 0.9909

c) 77  ± 3 * 5.5 = 60.5, 93.5

P(60.5 < x < 93.5) = P[(60.5 - 77)/ 5.5) < (x - ) /  < (93.5 - 77) / 5.5) ]

= P(-3.0 < z < 3.0)

= P(z < 3.0) - P(z < -3.0)

Using z table,

=0.9987 - 0.0013

= 0.9974

The percentage is = 99.74%