Chapter 7.3, Problem 33P

Estimating the Standard Deviation: Veterinary Science The resting heart rate for an adult horse should average about $\mu =46$ beats per minute with a ( $95\%$ of data) range from 22 to 70 beats per minute, based on information from the Merck Veterinary Manual (a classic reference used in most veterinary colleges). Let x be a random variable that represents the resting heart rate for an adult horse. Assume that x has a distribution that is approximately normal.

(a) Estimate the standard deviation of the x distribution. Hint: See Problem 31.

(b) What is the probability that the heart rate is fewer than 25 beats per minute?

(c) What is the probability that the heart rate is greater than 60 beats per minute?

(d) What is the probability that the heart rate is between 25 and 60 beats per minute?

(e) Inverse Normal Distribution A horse whose resting heart rate is in the upper 10% of the probability distribution of heart rates may have a secondary infection or illness that needs to be treated. What is the heart rate corresponding to the upper 10% cutoff point of the probability distribution?

To determine

The standard deviation of x values using rule of thumb.

Answer to Problem 33P

Solution:

The standard deviation of x values using rule of thumb is 12.0.

Explanation of Solution

According to the rule of thumb for estimating the standard deviation from 95% range of data values is:

$\begin{array}{l}\text{Standard}\text{deviation}\approx \frac{\text{Range}}{4}\\ \text{Standard}\text{deviation}\approx \frac{\text{High}\text{value}-\text{Low}\text{value}}{4}\\ \text{Standard}\text{deviation}\approx \frac{70-22}{4}\\ \text{Standard}\text{deviation}\approx 12\end{array}$

The standard deviation of x values using rule of thumb is 12.0.

To determine

The probability that heart rate is fewer than 25 beats per minute.

Answer to Problem 33P

Solution:

The probability that heart rate is fewer than 25 beats per minute is 0.0401.

Explanation of Solution

We have normal distribution with $\mu =46,S.D\approx 12$

$x=25$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{25-46}{12}=-1.75\\ P(x<25)=P(z<-1.75)\end{array}$

By using Table 3 from appendix

$P(x<25)=0.0401$

The probability that heart rate is fewer than 25 beats per minute is 0.0401.

To determine

The probability that heart rate is greater than 60 beats per minute.

Answer to Problem 33P

Solution:

The probability that heart rate is greater than 60 beats per minute is 0.121.

Explanation of Solution

We have normal distribution with $\mu =46,S.D\approx 12$

$x=60$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{60-46}{12}=1.17\\ P(x>60)=P(z>1.17)\\ P(x>60)=1-P(z\le 1.17)\end{array}$

By using Table 3 from appendix

$\begin{array}{l}P(x>60)=1-0.8790\\ P(x>60)=0.121\end{array}$

The probability that heart rate is greater than 60 beats per minute is 0.121.

To determine

The probability that heart rate is between 25 and 60 beats per minute.

Answer to Problem 33P

Solution:

The probability that heart rate is between 25 and 60 beats per minute is 0.8389.

Explanation of Solution

We have normal distribution with $\mu =46,S.D\approx 12$

$x_1=25,x_2=60$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z_1=\frac{25-46}{12}=-1.75\\ z_2=\frac{60-46}{12}=1.17\\ P(25<x<60)=P(-1.75<z<1.17)\\ P(25<x<60)=P(z<1.17)-P(z<-1.75)\end{array}$

By using Table 3 from appendix

$\begin{array}{l}P(25<x<60)=0.8790-0.0401\\ P(25<x<60)=0.8389\end{array}$

The probability that heart rate is between 25 and 60 beats per minute is 0.8389.

To determine

The heart rate corresponding to the upper 10% cutoff point of the probability distribution.

Answer to Problem 33P

Solution:

The heart rate corresponding to the upper 10% cutoff point of the probability distribution is 61 beats per minute.

Explanation of Solution

We have normal distribution with $\mu =46,S.D\approx 12$

A horse whose resting heart rate is in the upper 10% of the probability distribution of heart rates may have secondary infection or illness that needs to be treated.

From above we have $A=0.10$

Area left of z = 1 – A = 0.90

By using Table 3 from appendix

$z=1.285$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ x=z\sigma +\mu \\ x=1.285(12)+46\\ x=61.42\\ x\approx 61\end{array}$

The heart rate corresponding to the upper 10% cutoff point of the probability distribution is 61 beats per minute.