Chapter 7.3, Problem 34P

Estimating the Standard Deviation: Veterinary Science How much should a healthy kitten weigh? A healthy 10-week-old (domestic) kitten should weigh an average of $\mu =24.5$ ounces with a (95% of data) range from 14 to 35 ounces. (See reference in Problem 33.) Let x be a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten. 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 a healthy 10-week-old kitten will weigh less than 14 ounces?

(c) What is the probability that a healthy 10-week-old kitten will weigh more than 33 ounces?

(d) What is the probability that a healthy 10-week-old kitten will weigh between 14 and 33 ounces?

(e) Inverse Normal Distribution A kitten whose weight is in the bottom 10% of the probability distribution of weights is called undernourished. What is the cutoff point for the weight of an undernourished kitten?

To determine

The standard deviation of x values using rule of thumb.

Answer to Problem 34P

Solution:

The standard deviation of x values using rule of thumb is 5.25.

Explanation of Solution

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

$\begin{array}{l}\text{Standard}\text{deviation}\approx \frac{\text{Range}}{4}\\ \text{Standard}\text{deviation}\approx \frac{\text{High}\text{value}-\text{Low}\text{value}}{4}\\ \text{Standard}\text{deviation}\approx \frac{35-14}{4}\\ \text{Standard}\text{deviation}\approx 5.25\end{array}$

The standard deviation of x values using rule of thumb is 5.25.

To determine

The probability that a healthy 10 week old kitten will weigh less than 14 ounces.

Answer to Problem 34P

Solution:

The probability that a healthy 10 week old kitten will weigh less than 14 ounces is 0.0228.

Explanation of Solution

We have normal distribution with $\mu =24.5,S.D\approx 5.25$

$x=14$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{14-24.5}{5.25}=-2.0\\ P(x<14)=P(z<-2.0)\end{array}$

By using Table 3 from appendix

$P(x<14)=0.0228$

The probability that a healthy 10 week old kitten will weigh less than 14 ounces is 0.0228.

To determine

The probability that a healthy 10 week old kitten will weigh more than 33 ounces.

Answer to Problem 34P

Solution:

The probability that a healthy 10 week old kitten will weigh more than 33 ounces is 0.0526.

Explanation of Solution

We have normal distribution with $\mu =24.5,S.D\approx 5.25$

$x=33$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z=\frac{33-24.5}{5.25}=1.62\\ P(x>33)=P(z>1.62)\\ P(x>33)=1-P(z\le 1.62)\end{array}$

By using Table 3 from appendix

$\begin{array}{l}P(x>33)=1-0.9474\\ P(x>33)=0.0526\end{array}$

The probability that a healthy 10 week old kitten will weigh more than 33 ounces is 0.0526.

To determine

The probability that a healthy 10 week old kitten will weigh between 14 and 33 ounces.

Answer to Problem 34P

Solution:

The probability that a healthy 10 week old kitten will weigh between 14 and 33 ounces is 0.9246.

Explanation of Solution

We have normal distribution with $\mu =24.5,S.D\approx 5.25$

$x_1=14,x_2=33$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ z_1=\frac{14-24.5}{5.25}=-2.0\\ z_2=\frac{33-24.5}{5.25}=1.62\\ P(14<x<33)=P(-2.0<z<1.62)\\ P(14<x<33)=P(z<1.62)-P(z<-2.0)\end{array}$

By using Table 3 from appendix

$\begin{array}{l}P(14<x<33)=0.9474-0.0228\\ P(14<x<33)=0.9246\end{array}$

The probability that a healthy 10 week old kitten will weigh between 14 and 33 ounces is 0.9246.

To determine

The cutoff point for the weight of an undernourished kitten.

Answer to Problem 34P

Solution:

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.

Explanation of Solution

We have normal distribution with $\mu =24.5,S.D\approx 5.25$

A kitten whose weight is in the bottom 10% of the probability distribution of weights is called undernourished.

From above we have $A=0.10$

Area left of z = 0.10

By using Table 3 from appendix

$z=-1.285$

By using formula for normal distribution:-

$\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ x=z\sigma +\mu \\ x=-1.285(5.25)+24.5\\ x=17.75\\ x\approx 17.8\end{array}$

The cutoff point for the weight of an undernourished kitten is 17.8 ounces.