Chapter 7.2, Problem 11P

z Scores: Red Blood Cell Count Let x = red blood cell (RBC) count in millions per cubic millimeter of whole blood. For healthy females, x has an approximately normal distribution with mean $\mu =4.8$ and standard deviation $\sigma =0.3$ (based on information from Diagnostic Tests with Nursing Implications, edited by S. Loeb, Springhouse Press). Convert each of the following x intervals to z intervals.

$\left(\text{a}\right)4.5<x\left(\text{b}\right)x<4.2\left(\text{c}\right)4.0<x<5.5$

Convert each of the following z. intervals to x intervals.

$\left(\text{d}\right)z<-1.44\left(\text{e}\right)1.28<z \left(\text{f}\right)-2.25<z<-1.00$

(g) Interpretation If a female had an RBC count of 5.9 or higher, would that be considered unusually high? Explain using z values and Figure 7-12.

To determine

The z intervals from x interval $4.5<x$.

Answer to Problem 11P

Solution: The z intervals from x interval is $(-1.0<z)$.

Explanation of Solution

$\begin{array}{l}\mu =4.8,\sigma =0.3\\ x=4.5\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{4.5-4.8}{0.3}\\ \\ z=-1.0\end{array}$

$P(4.5<x)=P(-1.0<z)$

To determine

The z intervals from x interval $x<4.2$.

Answer to Problem 11P

Solution: The z intervals from x interval is $z<-2.0$.

Explanation of Solution

$\begin{array}{l}\mu =4.8,\sigma =0.3\\ x=4.2\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{4.2-4.8}{0.3}\\ \\ z=-2.0\end{array}$

$P(x<4.2)=P(z<-2.0)$

To determine

The z intervals from x interval $4.0<x<5.5$.

Answer to Problem 11P

Solution: The z intervals from x interval is $(-2.67<z<2.33)$.

Explanation of Solution

$\begin{array}{l}\mu =4.8,\sigma =0.3\\ x_1=4.0,x_2=5.5\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z_1=\frac{4.0-4.8}{0.3}=-2.67\\ \\ z_2=\frac{5.5-4.8}{0.3}=2.33\\ \end{array}$

$P(4.0<x<5.5)=P(-2.67<z<2.33)$

To determine

The x intervals from z interval $z<-1.44$.

Answer to Problem 11P

Solution: The x intervals from z interval is $x<4.37$.

Explanation of Solution

$\begin{array}{l}\mu =4.8,\sigma =0.3\\ z=-1.44\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \\ x=-1.44(0.3)+4.8\\ \\ x=4.37\end{array}$

$P(z<-1.44)=P(x<4.37)$

To determine

The x intervals from z interval $1.28<z$.

Answer to Problem 11P

Solution: The x intervals from z interval is $5.18<x$.

Explanation of Solution

$\begin{array}{l}\mu =4.8,\sigma =0.3\\ z=1.28\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \\ x=1.28(0.3)+4.8\\ \\ x=5.18\end{array}$

$P(1.28<z)=P(5.18<x)$

To determine

The x intervals from z interval $-2.25<z<-1.00$.

Answer to Problem 11P

Solution: The x intervals from z interval is $4.13<x<4.50$.

Explanation of Solution

$\begin{array}{l}\mu =4.8,\sigma =0.3\\ z_1=-2.25,z_2=-1.0\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \\ x_1=-2.25(0.3)+4.8\\ \\ x_1=4.13\\ \\ x_2=-1.0(0.3)+4.8\\ \\ x_2=4.50\end{array}$

$P(-2.25<z<-1.0)=P(4.13<x<4.50)$

To determine

Whether a female having a RBC count of 5.9 or higher would be considered unusually high.

Answer to Problem 11P

Solution: Yes, female having a RBC count of 5.9 or higher would be considered unusually high.

Explanation of Solution

$\begin{array}{l}\mu =4.8,\sigma =0.30\\ x=5.9\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{5.9-4.8}{0.30}\\ \\ z=3.67\end{array}$

According to Figure 7-12, 99.7% of the data values lie within 3 standard deviation of the means. Since the obtained z-value is 3.67 is above three standard deviation of means, hence we can say that female having a RBC count of 5.9 or higher would be considered unusually high.