The extract of a plant native to Taiwan has been tested as a possible treatment

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 63 and standard deviation of 10 grams per mililiter.

a) What is the probability that the amount of collagen is greater than 61 grams per mililiter?

b) What is the probability that the amount of collagen is less than 86 grams per mililiter?

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

Explanation / Answer

mean = 63 , s = 10

a) P( x > 61)

z = ( x - mean) / s
= ( 61 - 63) / 10
= -0.2

we need to find P(z > -0.2)
P (x > 61) = P(z > -0.2) = 0.5793

b) 68 of

P( x < 86)

z = ( x - mean) / s
= ( 86 - 63) / 10
= 2.3

we need to find P(z < 2.3)
P (x < 86) = P(z < 2.3) = 0.9893

c)

In order to find the extract of the plant that fall within 1 s.d. of the mean, we need to find the z-value respective to 1 s.d. i.e. +/-1

Hence P(-1 < z < 1) = P(z < 1) - P(z < -1)
Using standard normal table, we have
P(z < 1) = 0.8413
P(z < -1) = 0.1587

P(-1 < z < 1) = 0.8413 - 0.1587 = 0.6827

Hence 68.27%

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