Question

listed below in order are prices in dollars for a big mac hamburger in the united states, canada, mexico, china, japan, russia, switzerland, italy, spain, britain, india, and egypt. such data are used to compare currency exchange rates and the costs of goods in different countries. find the range, variance, and standard deviation for the given sample data. what do the measures of variation tell us about the prices of a big mac in different countries? the range is 4.9 dollars. the variance is (type an integer or decimal rounded to two decimal places as needed.) dollars²? (type an integer or decimal rounded to two decimal places as needed.) dollars. the data points are: 5.31, 5.29, 2.62, 3.22, 3.39, 2.29, 6.81, 5.06, 4.78, 4.39, 2.76, 1.91.

Explanation:

Step1: List the data points

The data points are: \(5.31, 5.29, 2.62, 3.22, 3.39, 2.29, 6.81, 5.06, 4.78, 4.39, 2.76, 1.91\)

Step2: Find the mean (\(\bar{x}\))

First, sum all the data points:
\[

$$\begin{align*} 5.31 + 5.29 + 2.62 + 3.22 + 3.39 + 2.29 + 6.81 + 5.06 + 4.78 + 4.39 + 2.76 + 1.91&=(5.31 + 5.29)+(2.62 + 3.22)+(3.39 + 2.29)+(6.81 + 5.06)+(4.78 + 4.39)+(2.76 + 1.91)\\ &=10.6 + 5.84 + 5.68 + 11.87 + 9.17 + 4.67\\ &=10.6+5.84 = 16.44\\ 16.44+5.68 = 22.12\\ 22.12+11.87 = 33.99\\ 33.99+9.17 = 43.16\\ 43.16+4.67 = 47.83 \end{align*}$$

\]
There are \(n = 12\) data points. So the mean \(\bar{x}=\frac{47.83}{12}\approx3.9858\)

Step3: Calculate the squared differences from the mean

For each data point \(x_i\), calculate \((x_i - \bar{x})^2\):

• \((5.31 - 3.9858)^2=(1.3242)^2\approx1.7535\)
• \((5.29 - 3.9858)^2=(1.3042)^2\approx1.6990\)
• \((2.62 - 3.9858)^2=(-1.3658)^2\approx1.8654\)
• \((3.22 - 3.9858)^2=(-0.7658)^2\approx0.5864\)
• \((3.39 - 3.9858)^2=(-0.5958)^2\approx0.3548\)
• \((2.29 - 3.9858)^2=(-1.6958)^2\approx2.8757\)
• \((6.81 - 3.9858)^2=(2.8242)^2\approx7.9761\)
• \((5.06 - 3.9858)^2=(1.0742)^2\approx1.1539\)
• \((4.78 - 3.9858)^2=(0.7942)^2\approx0.6308\)
• \((4.39 - 3.9858)^2=(0.4042)^2\approx0.1634\)
• \((2.76 - 3.9858)^2=(-1.2258)^2\approx1.5026\)
• \((1.91 - 3.9858)^2=(-2.0758)^2\approx4.3090\)

Step4: Sum the squared differences

\[

$$\begin{align*} &1.7535 + 1.6990 + 1.8654 + 0.5864 + 0.3548 + 2.8757 + 7.9761 + 1.1539 + 0.6308 + 0.1634 + 1.5026 + 4.3090\\ &=(1.7535+1.6990)+(1.8654+0.5864)+(0.3548+2.8757)+(7.9761+1.1539)+(0.6308+0.1634)+(1.5026+4.3090)\\ &=3.4525 + 2.4518 + 3.2305 + 9.13 + 0.7942 + 5.8116\\ &=3.4525+2.4518 = 5.9043\\ &5.9043+3.2305 = 9.1348\\ &9.1348+9.13 = 18.2648\\ &18.2648+0.7942 = 19.059\\ &19.059+5.8116 = 24.8706 \end{align*}$$

\]

Step5: Calculate the variance (\(s^2\))

For a sample, variance \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\)
Here, \(n=12\), so \(n - 1=11\)
\(s^2=\frac{24.8706}{11}\approx2.26096\approx2.26\) (rounded to two decimal places)

Answer:

The variance is \(\boxed{2.26}\) dollars\(^2\)