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 72 and standard deviation of 5.1 grams per mililiter.
(a) What is the probability that the amount of collagen is greater than 63 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?

Answer 1

Part a)

X ~ N ( µ = 72 , σ = 5.1 )
P ( X > 63 ) = 1 - P ( X < 63 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 63 - 72 ) / 5.1
Z = -1.7647
P ( ( X - µ ) / σ ) > ( 63 - 72 ) / 5.1 )
P ( Z > -1.7647 )
P ( X > 63 ) = 1 - P ( Z < -1.7647 )
P ( X > 63 ) = 1 - 0.0388
P ( X > 63 ) = 0.9612

Part b)

X ~ N ( µ = 72 , σ = 5.1 )
P ( X < 86 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 86 - 72 ) / 5.1
Z = 2.7451
P ( ( X - µ ) / σ ) < ( 86 - 72 ) / 5.1 )
P ( X < 86 ) = P ( Z < 2.7451 )
P ( X < 86 ) = 0.997

Part c)

Area within 1 standard deviation i.e Z score ± 1

X = µ ± Zσ = 72 ± ( 1 * 5.1 )

X1 = 66.9

X2 = 77.1

X ~ N ( µ = 72 , σ = 5.1 )
P ( 66.9 < X < 77.1 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 66.9 - 72 ) / 5.1
Z = -1
Z = ( 77.1 - 72 ) / 5.1
Z = 1
P ( -1 < Z < 1 )
P ( 66.9 < X < 77.1 ) = P ( Z < 1 ) - P ( Z < -1 )
P ( 66.9 < X < 77.1 ) = 0.8413 - 0.1587
P ( 66.9 < X < 77.1 ) = 0.6827

Percentage = 0.6827 * 100 = 68.27%