Question

A ferritin test is a popular test to measure a person's current iron stores. In women, ferritin is approximately normally distributed with a mean of 89 ng/mL and a standard deviation of 23 ng/mL. a.)What is the probability that a women has a ferritin value of 100 or less? b.) If a women has a ferritin of 140, what percentile is this? c.) If 50 women are tested, what is the probability that the mean ferritin exceeds 90?

Answer 1

Part a)
X ~ N ( µ = 89 , σ = 23 )
P ( X <= 100 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 100 - 89 ) / 23
Z = 0.4783
P ( ( X - µ ) / σ ) < ( 100 - 89 ) / 23 )
P ( X <= 100 ) = P ( Z < 0.4783 )
P ( X <= 100 ) = 0.6838

Part b)
X ~ N ( µ = 89 , σ = 23 )
P ( X < 140 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 140 - 89 ) / 23
Z = 2.2174
P ( ( X - µ ) / σ ) < ( 140 - 89 ) / 23 )
P ( X < 140 ) = P ( Z < 2.2174 )
P ( X < 140 ) = 0.9867

Percentile = 0.9867 * 100 = 98.67 ≈ 99^th percentile.

Part c)
X ~ N ( µ = 89 , σ = 23 )
P ( X > 90 ) = 1 - P ( X < 90 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 90 - 89 ) / ( 23 / √ ( 50 ) )
Z = 0.3074
P ( ( X - µ ) / ( σ / √ (n)) > ( 90 - 89 ) / ( 23 / √(50) )
P ( Z > 0.31 )
P ( X̅ > 90 ) = 1 - P ( Z < 0.31 )
P ( X̅ > 90 ) = 1 - 0.6207
P ( X̅ > 90 ) = 0.3793