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. find the value closest to the probability that a randomly selected woman has a red blood cell count above the normal range of 4.2 to 5.4 million cells per microliter. round to four decimal places.
a. 0.0156
b. 0.1611
c. 0.9842
d. 0.0409

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 \(x = 5.4\), \(z_1=\frac{5.4 - 4.577}{0.382}=\frac{0.823}{0.382}\approx2.15\).
For \(x = 4.2\), \(z_2=\frac{4.2 - 4.577}{0.382}=\frac{- 0.377}{0.382}\approx - 0.99\).

Step2: Find the probability \(P(4.2\lt X\lt5.4)\)

Using the standard normal distribution table, \(P(Z\lt2.15)=0.9842\), \(P(Z\lt - 0.99)=0.1611\).
Then \(P(4.2\lt X\lt5.4)=P(-0.99\lt Z\lt2.15)=P(Z\lt2.15)-P(Z\lt - 0.99)=0.9842-0.1611 = 0.8231\).

Step3: Find the probability \(P(X\gt5.4)\)

Since \(P(X\gt5.4)=1 - P(X\leqslant5.4)\), and \(P(X\leqslant5.4)=P(Z\leqslant2.15) = 0.9842\) (from the standard normal table). But we can also calculate it using the total probability.
We know that the total probability \(P = 1\). The probability that \(X\) is in the range \(4.2\) to \(5.4\) is \(P(4.2\lt X\lt5.4) = 0.8231\), and \(P(X\lt4.2)=P(Z\lt - 0.99)=0.1611\).
\(P(X\gt5.4)=1-(P(X\lt4.2)+P(4.2\lt X\lt5.4))=1-(0.1611 + 0.8231)=0.0158\approx0.0156\) (due to rounding differences in table - lookup).

Answer:

A. \(0.0156\)