A marketing expert wanted to see if a popular restaurant could survive in a new location. He was worried that even though the restaurant was popular, people would not want to travel out of their way just to go to it. The expert decided to survey five samples of 100 people throughout the city to see how many would say "Yes, I would go" to the restaurant in its new location. The results follow
41, 45, 50, 55, 64
For comparison purposes, the marketing expert took five additional samples of 100 people throughout the city to determine if people would dine at the same restaurant if it were right in the heart of the city. The mean number of people who said "Yes, I would go" was 75 with a standard deviation of 15 people.
(a)
What is the coefficient of variation for the number of people who said they would travel out of their way to the restaurant in its new location?

The **coefficient of variation (CV)** is a statistical measure that shows the extent of variability in relation to the mean of a data set. It is calculated as the ratio of the standard deviation to the mean, expressed as a percentage:

\[
\text{CV} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100
\]

### Step 1: Calculate the Mean

We are given the following data for the number of people who said "Yes, I would go" to the restaurant in its new location: 

\[
41, 45, 50, 55, 64
\]

The mean \(\mu\) is calculated as:

\[
\mu = \frac{41 + 45 + 50 + 55 + 64}{5} = \frac{255}{5} = 51
\]

### Step 2: Calculate the Standard Deviation

To calculate the standard deviation \(\sigma\), we first need to find the variance. The formula for the variance is:

\[
\sigma^2 = \frac{\sum (x_i - \mu)^2}{n}
\]

Where:
- \( x_i \) are the data points,
- \( \mu \) is the mean,
- \( n \) is the number of data points.

Now, calculate each squared difference from the mean:

\[
(41 - 51)^2 = (-10)^2 = 100
\]
\[
(45 - 51)^2 = (-6)^2 = 36
\]
\[
(50 - 51)^2 = (-1)^2 = 1
\]
\[
(55 - 51)^2 = (4)^2 = 16
\]
\[
(64 - 51)^2 = (13)^2 = 169
\]

The sum of the squared differences is:

\[
100 + 36 + 1 + 16 + 169 = 322
\]

The variance is:

\[
\sigma^2 = \frac{322}{5} = 64.4
\]

Now, take the square root to find the standard deviation:

\[
\sigma = \sqrt{64.4} \approx 8.02
\]

### Step 3: Calculate the Coefficient of Variation

Now that we have the mean (\( \mu = 51 \)) and the standard deviation (\( \sigma \approx 8.02 \)), we can calculate the coefficient of variation:

\[
\text{CV} = \frac{8.02}{51} \times 100 \approx 15.73\%
\]

Thus, the coefficient of variation for the number of people who said they would travel out of their way to the restaurant in its new location is approximately **15.73%**.

Would you like details on interpreting this result or have any questions?

---

Here are 5 related questions:
1. What does a high or low coefficient of variation imply in a data set?
2. How would the results differ if we used a different sample size for the survey?
3. How does the coefficient of variation compare with other measures of variability?
4. What factors could influence people's willingness to travel to a new restaurant location?
5. How can businesses use the coefficient of variation in decision-making?

**Tip:** The coefficient of variation is particularly useful when comparing the relative variability of data sets with different units or means.

A marketing expert wanted to see if a popular restaurant could survive in a new location. He was worried that even though the restaurant was popular, people would not want to travel out of their way just to go to it. The expert decided to survey five samples of 100 people throughout the city to see how many would say 'Yes, I would go' to the restaurant in its new location. The results follow 41, 45, 50, 55, 64. For comparison purposes, the marketing expert took five additional samples of 100 people throughout the city to determine if people would dine at the same restaurant if it were right in the heart of the city. The mean number of people who said 'Yes, I would go' was 75 with a standard deviation of 15 people. (a) What is the coefficient of variation for the number of people who said they would travel out of their way to the restaurant in its new location?

Learn how to calculate the coefficient of variation for a marketing survey that asked whether people would travel to a new restaurant location. Using survey data points (41, 45, 50, 55, 64), this problem walks through the steps to find the mean, standard deviation, and CV. Suitable for Grades 10-12.

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