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QUAN 2010 Notes: Introduction to Business Statistics

Omar Ortega

Section 3.3 Using the Mean and Standard Deviation Together

There are a few important concepts that use the mean and standard deviation together to describe data sets.

Definition 3.3.1.

The coefficient of variation (cv) measures the standard deviation in terms of its percentage of the mean.
\begin{equation*} \text{formula:} \text{ cv}=\frac{\sigma}{\mu}\cdot 100\text{ or } \frac{s}{\overline{x}}\cdot 100 \end{equation*}

Definition 3.3.2.

The \(z\)-score identifies the number of standard deviations a particular value is from the mean of its distribution.
\begin{equation*} \text{formula: } z=\frac{x-\mu}{\sigma}\text{ or } \frac{x-\overline{x}}{s} \end{equation*}
\begin{equation*} \text{Excel formula: STANDARDIZE} \end{equation*}

Exercise 3.3.3.

The mean weight of newborn infants is 7 pounds, and the standard deviation is 0.8 pound. The weights of newborn infants are normally distributed. (This is a symmetric distribution that we’ll talk more about in a later chapter.)
Find the \(z\)-scores for the following weights.

(a)

9 pounds
Answer.
\begin{equation*} z_9=\frac{9-7}{0.8}=2.5 \end{equation*}

(b)

7 pounds
Answer.
\begin{equation*} z_7=\frac{7-7}{0.8}=0 \end{equation*}

(c)

6 pounds
Answer.
\begin{equation*} z_6=\frac{6-7}{0.8}=-1.25 \end{equation*}

Question 3.3.4.

What do the \(z\)-scores in Exercise 3.3.3 mean about how the weights compare to the average weight of newborn infants?

Exercise 3.3.5.

An employee satisfaction survey was given to employees, and they were asked to rate their overall satisfaction with their job from 0 to 10 (inclusive). The average rating was 5.5, and the standard deviation was 1.2. What is the satisfaction rating corresponding to a \(z\)-score of \(-1.5\text{?}\)
Answer.
\begin{equation*} \text{satisfaction rating}=\text{mean}-1.5(\text{standard deviation})=5.5-1.5(1.2)=5.5-1.8=3.7 \end{equation*}

Subsection 3.3.1 The Empirical Rule

The Empirical Rule says that if a distribution follows a bell-shaped, symmetric curve centered around the mean, then we should expect:
  • approximately \(68\%\) of the data to fall within one standard deviation of the mean,
  • approximately \(95\%\) of the data to fall within two standard deviations of the mean, and
  • approximately \(99.7\%\) of the data to fall within three standard deviations of the mean.
Figure 3.3.6.
The Empirical Rule (Made in GeoGebra by Kady Schneiter)

Exercise 3.3.7. Empirical Rule.

Exercise 3.3.8. Empirical Rule.

Exercise 3.3.9. Empirical Rule.

Exercise 3.3.11.

Assume the average selling price for houses in a certain county is \(\$ 348,000\) with a standard deviation of \(\$ 30,000\text{.}\)
(a)
Determine the coefficient of variation.
Answer.
\begin{equation*} CV = \frac{\sigma}{\mu}\cdot 100=\frac{30000}{348000}\cdot 100 = 8.62\% \end{equation*}
(b)
Calculate the \(z\)-score for a house that sells for \(\$ 379,000\text{.}\)
Answer.
\begin{equation*} z=\frac{x-\mu}{\sigma}=\frac{379000-348000}{30000}=1.03 \end{equation*}
(c)
Using the Empirical Rule, determine the range of prices that includes \(68\%\) of the homes around the mean. (To do this, we need to assume that the selling prices of homes is bell-shaped and symmetric.)
Answer.
\(68\%\) of prices are within 1 standard deviation of the mean. That is, \(68\%\) of prices are between the numbers \(\mu\pm\sigma\) (assuming the distribution of prices is bell-shaped and symmetric).
\begin{equation*} 348000\pm 30000\rightarrow \boxed{\$ 318,000 \text{ to } \$378,000} \end{equation*}
(d)
Using Chebyshev’s Theorem, determine the range of prices that includes at least \(91\%\) of the homes around the mean. (Using Chebyshev’s Theorem, we don’t need to assume anything about how the prices are distributed.)
Answer.
\begin{equation*} 1-\frac{1}{z^2}=0.91 \end{equation*}
\begin{equation*} 1-0.91=\frac{1}{z^2} \end{equation*}
\begin{equation*} 0.09 = \frac{1}{z^2} \end{equation*}
\begin{equation*} 0.09z^2 = 1 \end{equation*}
\begin{equation*} z^2=\frac{1}{0.09} \end{equation*}
\begin{equation*} z=\sqrt{\frac{1}{0.09}} \end{equation*}
So the range of prices including values within \(\sqrt{\frac{1}{0.09}}\) of the mean will include at least \(91\%\) of the homes.
\begin{equation*} \mu\pm \sqrt{\frac{1}{0.09}}\sigma \end{equation*}
\begin{equation*} 348,000\pm \sqrt{\frac{1}{0.09}}\cdot 30,000 \end{equation*}
\begin{equation*} \boxed{\$248,000\text{ to }\$ 448,000} \end{equation*}

Exercise 3.3.12.

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