Chapter 7.2, Problem 10P

z Scores: Fawns Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that is approximately normally distributed with mean $\mu =27.2$ kilograms and standard deviation $\sigma =4.3$ kilograms (based on information from The Mule Deer of Mesa Verde National Park, by G. W. Mierau and J. L. Schmidt, Mesa Verde Museum Association). Let x be the weight of a fawn in kilograms. Convert each of the following x intervals to z intervals.

Convert each of the following z intervals to x intervals.

$\left(\text{d}\right)-2.17<z\left(\text{e}\right)z<1.28\left(\text{f}\right)-1.99<z<1.44$

(g) Interpretation If a fawn weighs 14 kilograms, would you say it is an unusually small animal? Explain using z values and Figure 7-12.

(h) Interpretation If a fawn is unusually large, would you say that the z value for the weight of the fawn will be close to $0,–2,\text{ or }3?$ Explain.

To determine

The z intervals from x interval $x<30$.

Answer to Problem 10P

Solution: The z intervals from x interval is $z<0.65$.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ x=30\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{30-27.2}{4.3}\\ \\ z=0.65\end{array}$

$P(x<30)=P(z<0.65)$

To determine

The z intervals from x interval $19<x$.

Answer to Problem 10P

Solution: The z intervals from x interval is $-1.91<z$.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ x=19\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{19-27.2}{4.3}\\ \\ z=-1.91\end{array}$

$P(19<x)=P(-1.91<z)$

To determine

The z intervals from x interval $32<x<35$.

Answer to Problem 10P

Solution: The z intervals from x interval is $(1.12<z<1.81)$.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ x_1=32,x_2=35\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z_1=\frac{32-27.2}{4.3}=1.12\\ \\ z_2=\frac{35-27.2}{4.3}=1.81\\ \end{array}$

$P(32<x<35)=P(1.12<z<1.81)$

To determine

The x intervals from z interval $-2.17<z$.

Answer to Problem 10P

Solution: The x intervals from z interval is $17.87<x$.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ z=-2.17\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \\ x=-2.17(4.3)+27.2\\ \\ x=17.87\end{array}$

$P(-2.17<z)=P(17.87<x)$

To determine

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

Answer to Problem 10P

Solution: The x intervals from z interval is $x<32.70$.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ z=1.28\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \\ x=1.28(4.3)+27.2\\ \\ x=32.70\end{array}$

$P(z<1.28)=P(x<32.70)$

To determine

The x intervals from z interval $-1.99<z<1.44$.

Answer to Problem 10P

Solution: The x intervals from z interval is $18.64<x<33.39$.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ z_1=-1.99,z_2=1.44\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \\ x_1=-1.99(4.3)+27.2\\ \\ x_1=18.64\\ \\ x_2=1.44(4.3)+27.2\\ \\ x_2=33.39\end{array}$

$P(-1.99<z<1.44)=P(18.64<x<33.39)$

To determine

Whether a fawn weighing 14 kilograms would be unusually small animal.

Answer to Problem 10P

Solution: Yes, we can say that it is an unusually small animal.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ x=14\end{array}$

We use the formula for normal distribution:

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{14-27.2}{4.3}\\ \\ z=-3.07\end{array}$

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 -3.07 is below three standard deviation of mean, hence we can say that fawn weighing 14 kilograms would be unusually small animal.

To determine

The z value for a fawn to be unusually large.

Answer to Problem 10P

Solution: The z value for a fawn to be unusually large is 3.

Explanation of Solution

$\begin{array}{l}\mu =27.2,\sigma =4.3\\ z_1=0,z_1=-2,z_1=3\end{array}$

$z=-2$ would be approximately 2 standard deviation below the mean, $z=0$ would be approximately equal to the mean and $z=3$ would be approximately 3 standard deviation above the mean which would be a unusually large value, hence z value for a fawn to be unusually large is 3.