Math Problem Statement

The fuel efficiency, in miles per gallon, of 12 small utility trucks was measured. The results are recorded in the table below. Fuel Efficiency (mpg) 23 24 24 27 14 25 25 31 24 23 24 23 Find the mean and sample standard deviation of these data. Round to the nearest hundredth.

Solution

To calculate the mean and sample standard deviation for the given data:

1. Data:

Fuel Efficiency (mpg):
23, 24, 24, 27, 14, 25, 25, 31, 24, 23, 24, 23

2. Steps to Compute the Mean:

The mean is given by:

xˉ=xin\bar{x} = \frac{\sum x_i}{n}

Where xi\sum x_i is the sum of all data points, and nn is the number of data points.

xi=23+24+24+27+14+25+25+31+24+23+24+23=267\sum x_i = 23 + 24 + 24 + 27 + 14 + 25 + 25 + 31 + 24 + 23 + 24 + 23 = 267 n=12n = 12

xˉ=2671222.25(rounded to the nearest hundredth).\bar{x} = \frac{267}{12} \approx 22.25 \, \text{(rounded to the nearest hundredth)}.

3. Steps to Compute the Sample Standard Deviation:

The sample standard deviation ss is given by:

s=(xixˉ)2n1s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}

Step 1: Compute Deviations from the Mean (xixˉx_i - \bar{x}):

xˉ=22.25\bar{x} = 22.25

&23 - 22.25 = 0.75, \quad 24 - 22.25 = 1.75, \quad 24 - 22.25 = 1.75, \\ &27 - 22.25 = 4.75, \quad 14 - 22.25 = -8.25, \quad 25 - 22.25 = 2.75, \\ &25 - 22.25 = 2.75, \quad 31 - 22.25 = 8.75, \quad 24 - 22.25 = 1.75, \\ &23 - 22.25 = 0.75, \quad 24 - 22.25 = 1.75, \quad 23 - 22.25 = 0.75. \end{aligned}$$ #### Step 2: Square Each Deviation: $$\begin{aligned} &0.75^2 = 0.5625, \quad 1.75^2 = 3.0625, \quad 1.75^2 = 3.0625, \\ &4.75^2 = 22.5625, \quad (-8.25)^2 = 68.0625, \quad 2.75^2 = 7.5625, \\ &2.75^2 = 7.5625, \quad 8.75^2 = 76.5625, \quad 1.75^2 = 3.0625, \\ &0.75^2 = 0.5625, \quad 1.75^2 = 3.0625, \quad 0.75^2 = 0.5625. \end{aligned}$$ $$\sum (x_i - \bar{x})^2 = 196.625$$ #### Step 3: Compute the Variance: $$s^2 = \frac{196.625}{12 - 1} = \frac{196.625}{11} \approx 17.875$$ #### Step 4: Compute the Standard Deviation: $$s = \sqrt{17.875} \approx 4.23 \, \text{(rounded to the nearest hundredth)}.$$ ### Final Results: - **Mean**: $$\bar{x} \approx 22.25 \, \text{mpg}$$ - **Sample Standard Deviation**: $$s \approx 4.23 \, \text{mpg}$$ Would you like additional clarification or details? 😊 --- #### Related Questions: 1. How is the sample standard deviation different from the population standard deviation? 2. Why is the denominator $$n-1$$ used in the sample standard deviation formula? 3. Can the mean and standard deviation be affected by outliers? If so, how? 4. How would you compute the variance for this dataset? 5. What does the standard deviation tell us about the spread of the data? --- #### Tip: When calculating deviations, keep intermediate steps precise to avoid rounding errors, and round only in the final results.

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Math Problem Analysis

Mathematical Concepts

Descriptive Statistics
Mean
Standard Deviation

Formulas

Mean: \( \bar{x} = \frac{\sum x_i}{n} \)
Sample Standard Deviation: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \)

Theorems

Properties of Variance and Standard Deviation

Suitable Grade Level

Grades 10-12