QUESTION IMAGE
Question
- you are at the department of motor vehicles waiting to get your drivers license. the wait times (in minutes) of the people in front of you have been 11, 7, 14, 2, 8, 13, 3, 6, 10, 3, 8, 4, 8, 4, 7
mean \tmedian \tmode \trange \tstandard deviation
- the data set gives the quiz scores for a student on quizzes consisting of 10 questions each: 7, 9, 7, 10, 8, 7, 9
mean \tmedian \tmode \trange \tstandard deviation
- the list shows the average price of a gallon of gasoline each year from 1994 to 2004: $1.04, $1.13, $1.13, $1.26, $1.13, $.97, $1.30, $1.47, $1.14, $1.47, $1.59
mean \tmedian \tmode \trange \tstandard deviation
Problem 3
Step1: Organize the data
The wait times are: \( 11, 7, 14, 2, 8, 13, 3, 6, 10, 3, 8, 4, 8, 4, 7 \). First, sort them: \( 2, 3, 3, 4, 4, 6, 7, 7, 8, 8, 8, 10, 11, 13, 14 \). Number of data points \( n = 15 \).
Step2: Calculate the Mean
Mean \( \bar{x}=\frac{\sum x}{n} \). Sum of data: \( 2 + 3 + 3 + 4 + 4 + 6 + 7 + 7 + 8 + 8 + 8 + 10 + 11 + 13 + 14 = 108 \). So, \( \bar{x}=\frac{108}{15}=7.2 \).
Step3: Calculate the Median
Median is the middle value. For \( n = 15 \) (odd), the 8th value. Sorted data: 8th value is \( 7 \).
Step4: Determine the Mode
Mode is the most frequent value. \( 8 \) appears 3 times, more than others.
Step5: Calculate the Range
Range = Maximum - Minimum = \( 14 - 2 = 12 \).
Step6: Calculate the Standard Deviation
First, find \( \sum (x - \bar{x})^2 \):
- For \( x = 2 \): \( (2 - 7.2)^2 = 27.04 \)
- \( x = 3 \): \( (3 - 7.2)^2 = 17.64 \) (two times)
- \( x = 4 \): \( (4 - 7.2)^2 = 10.24 \) (two times)
- \( x = 6 \): \( (6 - 7.2)^2 = 1.44 \)
- \( x = 7 \): \( (7 - 7.2)^2 = 0.04 \) (two times)
- \( x = 8 \): \( (8 - 7.2)^2 = 0.64 \) (three times)
- \( x = 10 \): \( (10 - 7.2)^2 = 7.84 \)
- \( x = 11 \): \( (11 - 7.2)^2 = 14.44 \)
- \( x = 13 \): \( (13 - 7.2)^2 = 33.64 \)
- \( x = 14 \): \( (14 - 7.2)^2 = 46.24 \)
Sum these up: \( 27.04 + 2\times17.64 + 2\times10.24 + 1.44 + 2\times0.04 + 3\times0.64 + 7.84 + 14.44 + 33.64 + 46.24 = 199.2 \).
Variance \( s^2=\frac{\sum (x - \bar{x})^2}{n - 1}=\frac{199.2}{14}\approx14.2286 \). Standard Deviation \( s=\sqrt{14.2286}\approx3.77 \).
Step1: Organize the data
Quiz scores: \( 7, 9, 7, 10, 8, 7, 9 \). Sort them: \( 7, 7, 7, 8, 9, 9, 10 \). Number of data points \( n = 7 \).
Step2: Calculate the Mean
Mean \( \bar{x}=\frac{\sum x}{n} \). Sum: \( 7 + 7 + 7 + 8 + 9 + 9 + 10 = 57 \). So, \( \bar{x}=\frac{57}{7}\approx8.14 \).
Step3: Calculate the Median
Median is the 4th value (since \( n = 7 \), odd). Sorted data: 4th value is \( 8 \).
Step4: Determine the Mode
Mode is the most frequent value. \( 7 \) appears 3 times, more than others.
Step5: Calculate the Range
Range = Maximum - Minimum = \( 10 - 7 = 3 \).
Step6: Calculate the Standard Deviation
First, find \( \sum (x - \bar{x})^2 \):
- \( x = 7 \): \( (7 - 8.14)^2 \approx1.30 \) (three times)
- \( x = 8 \): \( (8 - 8.14)^2 \approx0.02 \)
- \( x = 9 \): \( (9 - 8.14)^2 \approx0.74 \) (two times)
- \( x = 10 \): \( (10 - 8.14)^2 \approx3.46 \)
Sum: \( 3\times1.30 + 0.02 + 2\times0.74 + 3.46 = 3.9 + 0.02 + 1.48 + 3.46 = 8.86 \).
Variance \( s^2=\frac{8.86}{6}\approx1.4767 \). Standard Deviation \( s=\sqrt{1.4767}\approx1.215 \).
Step1: Organize the data
Gasoline prices: \( \$1.04, \$1.13, \$1.13, \$1.26, \$1.13, \$0.97, \$1.30, \$1.47, \$1.14, \$1.47, \$1.59 \). Sort them: \( 0.97, 1.04, 1.13, 1.13, 1.13, 1.14, 1.26, 1.30, 1.47, 1.47, 1.59 \). Number of data points \( n = 11 \).
Step2: Calculate the Mean
Mean \( \bar{x}=\frac{\sum x}{n} \). Sum: \( 0.97 + 1.04 + 1.13 + 1.13 + 1.13 + 1.14 + 1.26 + 1.30 + 1.47 + 1.47 + 1.59 = 13.23 \). So, \( \bar{x}=\frac{13.23}{11}\approx1.2027 \).
Step3: Calculate the Median
Median is the 6th value (since \( n = 11 \), odd). Sorted data: 6th value is \( 1.14 \).
Step4: Determine the Mode
Mode is the most frequent value. \( 1.13 \) appears 3 times, more than others.
Step5: Calculate the Range
Range = Maximum - Minimum = \( 1.59 - 0.97 = 0.62 \).
Step6: Calculate the Standard Deviation
First, find \( \sum (x - \bar{x})^2 \):
- \( x = 0.97 \): \( (0.97 - 1.2027)^2 \approx0.0542 \)
- \( x = 1.04 \): \( (1.04 - 1.2027)^2 \approx0.0265 \)
- \( x = 1.13 \): \( (1.13 - 1.2027)^2 \approx0.0053 \) (three times)
- \( x = 1.14 \): \( (1.14 - 1.2027)^2 \approx0.0039 \)
- \( x = 1.26 \): \( (1.26 - 1.2027)^2 \approx0.0033 \)
- \( x = 1.30 \): \( (1.30 - 1.2027)^2 \approx0.0095 \)
- \( x = 1.47 \): \( (1.47 - 1.2027)^2 \approx0.0715 \) (two times)
- \( x = 1.59 \): \( (1.59 - 1.2027)^2 \approx0.1500 \)
Sum: \( 0.0542 + 0.0265 + 3\times0.0053 + 0.0039 + 0.0033 + 0.0095 + 2\times0.0715 + 0.1500 \approx0.0542 + 0.0265 + 0.0159 + 0.0039 + 0.0033 + 0.0095 + 0.143 + 0.1500 = 0.4063 \).
Variance \( s^2=\frac{0.4063}{10}=0.04063 \). Standard Deviation \( s=\sqrt{0.04063}\approx0.2016 \).
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Mean: \( 7.2 \), Median: \( 7 \), Mode: \( 8 \), Range: \( 12 \), Standard Deviation: \( \approx3.77 \)