Question

In a typical blood sample, the surface areas of the red blood cells have a normal distribution with a mean of 135 µm2 and a standard deviation of 20 µm2 . What fraction of the red blood cells has a surface area larger than 170 µm2

Answer 1

Fraction of the red blood cells having a surface area larger than 170
`\mu m^(2)` is 0.04006 or 4% .

Explanation:

We are given that in a typical blood sample, the surface areas of the red blood cells have a normal distribution with a mean of 135
`\mu m^(2)` and a standard deviation of 20
`\mu m^(2)` .

Let X = the surface areas of the red blood cells, i.e;

X ~ N(
`\mu = 135,\sigma^(2) = 20^(2)`)

The Z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ Standard Normal(0,1)

So, fraction of the red blood cells having a surface area larger than 170
`\mu m^(2)` is given by P(X > 170
`\mu m^(2)` ) ;

P(X > 170) = P(
`(X-\mu)/(\sigma)` >
`(170-135)/(20)` ) = P(Z > 1.75) = 1 - P(Z <= 1.75)

= 1 - 0.95994 = 0.04006 or approx 4%.

Therefore, approx 4% of the red blood cells has a surface area larger than 170
`\mu m^(2)` .

Answer 2

Answer: P(x > 170) = 0.04

Explanation:

Since the the surface areas of the red blood cells in a typical blood sample have a normal distribution, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = surface areas of red blood cells.

µ = mean surface area

σ = standard deviation

From the information given,

µ = 135 µm2

σ = 20 µm2

We want to find the fraction of the red blood cells that has a surface area larger than 170 µm2. It is expressed as

P(x > 170) = 1 - P(x ≤ 170)

For x = 170,

z = (170 - 135)/20 = 1.75

Looking at the normal distribution table, the probability corresponding to the z score is 0.96

P(x > 170) = 1 - 0.96

P(x > 170) = 0.04