Question

assume that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per microliter and a standard deviation of 0.382 million cells per microliter. approximately what percentage of women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per microliter? round to two decimal places.

a. 82.26%
b. 16.11%
c. 4.09%
d. 17.69%

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 4.577\), \(\sigma=0.382\).
For \(x = 4.2\):
\(z_1=\frac{4.2 - 4.577}{0.382}=\frac{- 0.377}{0.382}\approx - 0.99\)
For \(x = 5.4\):
\(z_2=\frac{5.4 - 4.577}{0.382}=\frac{0.823}{0.382}\approx2.15\)

Step2: Use the standard normal distribution table

We want to find \(P(-0.99<Z<2.15)\).
We know that \(P(-0.99 < Z<2.15)=P(Z < 2.15)-P(Z<-0.99)\)
From the standard - normal table:
\(P(Z < 2.15)=0.9842\)
\(P(Z<-0.99) = 0.1611\)

Step3: Calculate the probability

\(P(-0.99 < Z<2.15)=0.9842-0.1611 = 0.8231\)

Answer:

A. \(82.26\%\)