Math / Data & Statistics

Question $\text{According to data from the city of Toronto, Ontario, Canada, there were more than 180,000 parking infractions in the city for December 2015, with fines totaling over 8,500,000 Canadian dollars. The fines (in Canadian dollars) for a random sample of 105 parking infractions in Toronto, Ontario, Canada, for December 2015 are listed below.}$
$\begin{array}{rrrrrrr} 30 & 30 & 30 & 30 & 40 & 60 & 40 \\ 15 & 50 & 150 & 40 & 30 & 30 & 30 \\ 40 & 30 & 40 & 30 & 30 & 30 & 40 \\ 40 & 40 & 30 & 60 & 60 & 30 & 150 \\ 40 & 30 & 250 & 40 & 30 & 30 & 30 \\ 30 & 30 & 30 & 40 & 30 & 40 & 30 \\ 50 & 15 & 40 & 40 & 30 & 40 & 30 \\ 40 & 30 & 30 & 40 & 30 & 30 & 30 \\ 100 & 30 & 40 & 30 & 30 & 30 & 40 \\ 30 & 30 & 30 & 40 & 100 & 30 & 40 \\ 30 & 40 & 30 & 40 & 40 & 40 & 40 \\ 30 & 30 & 30 & 60 & 30 & 40 & 40 \\ 30 & 40 & 15 & 60 & 30 & 15 & 150 \\ 150 & 40 & 40 & 30 & 30 & 150 & 60 \\ 30 & 40 & 60 & 30 & 40 & 40 & 30 \end{array}$
$\text{Parking Infractions by Time of Day}$ $\text{(Source: City of Toronto)}$
$\text{The figures above show parking infractions in Toronto, Ontario, Canada, for December 2015 by time of day and by day.}$
$\text{Hello! It looks like you've provided data on parking fines in Toronto for December 2015. However, I need more information to understand what specifically you need help with regarding this data. Could you please clarify the question or the problem you're trying to solve? For example, are you looking for a statistical analysis, a calculation of averages, or something else? Once I have a bit more context, I'll be happy to help!}$
$\text{Find the sample standard deviation, find the five number summary, make a frequency distribution, make a histogram}$

Studdy Solution

STEP 1

What is this asking? We're diving into the world of parking tickets in Toronto during December 2015!
We need to calculate the sample standard deviation and five-number summary for a given sample of fines, and then create a frequency distribution and histogram to visualize the data. Watch out! Don't mix up the sample standard deviation with the population standard deviation.
Also, remember that the five-number summary includes the minimum, first quartile (Q1), median (Q2), third quartile (Q3), and maximum.

STEP 2

1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Find the five-number summary.
4. Create a frequency distribution.
5. Create a histogram.

STEP 3

To kick things off, let's find the sample mean, which is just the average of all the fines in our sample.
We add up all the fines and then divide by the number of fines, which is $\text{105}$ .

STEP 4

$\text{Sum of fines} = 30 \cdot 32 + 40 \cdot 27 + 60 \cdot 8 + 15 \cdot 5 + 50 \cdot 2 + 100 \cdot 2 + 150 \cdot 3 + 250 = 960 + 1080 + 480 + 75 + 100 + 200 + 450 + 250 = 3595$

STEP 5

$\text{Sample Mean} = \frac{\text{Sum of fines}}{\text{Number of fines}} = \frac{3595}{105} \approx 34.24$ So, the sample mean is approximately $\$34.24$ .

STEP 6

Now, let's calculate the sample standard deviation, which tells us how spread out the fines are around the mean.
For each fine, we'll find the difference between the fine and the sample mean, square that difference, and add up all those squared differences.

STEP 7

$\text{Sum of squared differences} = 32(30-34.24)^2 + 27(40-34.24)^2 + 8(60-34.24)^2 + 5(15-34.24)^2 + 2(50-34.24)^2 + 2(100-34.24)^2 + 3(150-34.24)^2 + (250-34.24)^2 \approx 554.61 + 887.36 + 5279.04 + 1869.61 + 489.96 + 8554.36 + 41131.04 + 46849.61 = 106425.6$

STEP 8

Then, we divide that sum by the number of fines minus 1 (which is $\text{104}$ ) and take the square root of the result. $\text{Sample Standard Deviation} = \sqrt{\frac{\text{Sum of squared differences}}{\text{Number of fines} - 1}} = \sqrt{\frac{106425.6}{104}} \approx \sqrt{1023.32} \approx 31.99$ The sample standard deviation is approximately $\$31.99$ .

STEP 9

The five-number summary gives us a quick overview of the distribution of the fines.
It consists of the minimum, Q1, median (Q2), Q3, and maximum.

STEP 10

First, we arrange the data in ascending order.
The minimum is $\$15$ and the maximum is $\$250$ .

STEP 11

The median is the middle value.
Since we have $\text{105}$ data points, the median is the $53^{\text{rd}}$ value, which is $\$30$ .

STEP 12

Q1 is the median of the lower half of the data (excluding the median if the number of data points is odd).
In our case, Q1 is the $27^{\text{th}}$ value, which is $\$30$ .

STEP 13

Q3 is the median of the upper half of the data (excluding the median if the number of data points is odd).
Q3 is the $80^{\text{th}}$ value, which is $\$40$ .

STEP 14

So, our five-number summary is: $\text{Min} = \$15$ , $\text{Q1} = \$30$ , $\text{Median} = \$30$ , $\text{Q3} = \$40$ , $\text{Max} = \$250$ .

STEP 15

A frequency distribution organizes the data into classes (or intervals) and shows how many data points fall into each class.
We can create a table with the fine amounts and their corresponding frequencies.
For example, the fine of $\$30$ appears $\text{32}$ times.

STEP 16

A histogram is a visual representation of the frequency distribution.
It uses bars to show the frequency of each class.
The x-axis represents the fine amounts, and the y-axis represents the frequency.
The height of each bar corresponds to the frequency of the corresponding fine amount.

STEP 17

The sample standard deviation is approximately $\$31.99$ .
The five-number summary is: $\text{Min} = \$15$ , $\text{Q1} = \$30$ , $\text{Median} = \$30$ , $\text{Q3} = \$40$ , $\text{Max} = \$250$ .
The frequency distribution and histogram can be constructed as described above, showing the distribution of parking fines.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Find the five-number summary.
4. Create a frequency distribution.
5. Create a histogram.

STEP 3

To kick things off, let's find the sample mean, which is just the average of all the fines in our sample.
We add up all the fines and then divide by the number of fines, which is $\text{105}$ .

STEP 4

$\text{Sum of fines} = 30 \cdot 32 + 40 \cdot 27 + 60 \cdot 8 + 15 \cdot 5 + 50 \cdot 2 + 100 \cdot 2 + 150 \cdot 3 + 250 = 960 + 1080 + 480 + 75 + 100 + 200 + 450 + 250 = 3595$

STEP 5

$\text{Sample Mean} = \frac{\text{Sum of fines}}{\text{Number of fines}} = \frac{3595}{105} \approx 34.24$ So, the sample mean is approximately $\$34.24$ .

STEP 6

Now, let's calculate the sample standard deviation, which tells us how spread out the fines are around the mean.
For each fine, we'll find the difference between the fine and the sample mean, square that difference, and add up all those squared differences.

STEP 7

$\text{Sum of squared differences} = 32(30-34.24)^2 + 27(40-34.24)^2 + 8(60-34.24)^2 + 5(15-34.24)^2 + 2(50-34.24)^2 + 2(100-34.24)^2 + 3(150-34.24)^2 + (250-34.24)^2 \approx 554.61 + 887.36 + 5279.04 + 1869.61 + 489.96 + 8554.36 + 41131.04 + 46849.61 = 106425.6$

STEP 8

STEP 9

The five-number summary gives us a quick overview of the distribution of the fines.
It consists of the minimum, Q1, median (Q2), Q3, and maximum.

STEP 10

First, we arrange the data in ascending order.
The minimum is $\$15$ and the maximum is $\$250$ .

STEP 11

The median is the middle value.
Since we have $\text{105}$ data points, the median is the $53^{\text{rd}}$ value, which is $\$30$ .

STEP 12

Q1 is the median of the lower half of the data (excluding the median if the number of data points is odd).
In our case, Q1 is the $27^{\text{th}}$ value, which is $\$30$ .

STEP 13

Q3 is the median of the upper half of the data (excluding the median if the number of data points is odd).
Q3 is the $80^{\text{th}}$ value, which is $\$40$ .

STEP 14

So, our five-number summary is: $\text{Min} = \$15$ , $\text{Q1} = \$30$ , $\text{Median} = \$30$ , $\text{Q3} = \$40$ , $\text{Max} = \$250$ .

STEP 15

A frequency distribution organizes the data into classes (or intervals) and shows how many data points fall into each class.
We can create a table with the fine amounts and their corresponding frequencies.
For example, the fine of $\$30$ appears $\text{32}$ times.

STEP 16

A histogram is a visual representation of the frequency distribution.
It uses bars to show the frequency of each class.
The x-axis represents the fine amounts, and the y-axis represents the frequency.
The height of each bar corresponds to the frequency of the corresponding fine amount.

STEP 17

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Studdy Solution

STEP 1

STEP 2

1. Calculate the sample mean.2. Calculate the sample standard deviation.3. Find the five-number summary.4. Create a frequency distribution.5. Create a histogram.

STEP 3

To **kick things off**, let's find the sample mean, which is just the average of all the fines in our sample.We **add up** all the fines and then **divide** by the number of fines, which is 105\text{105}105.

STEP 4

STEP 5

Sample Mean=Sum of finesNumber of fines=3595105≈34.24\text{Sample Mean} = \frac{\text{Sum of fines}}{\text{Number of fines}} = \frac{3595}{105} \approx 34.24Sample Mean=Number of finesSum of fines​=1053595​≈34.24 So, the **sample mean** is approximately $34.24\$34.24$34.24.

STEP 6

Now, let's **calculate** the sample standard deviation, which tells us how spread out the fines are around the mean.For each fine, we'll find the difference between the fine and the sample mean, square that difference, and add up all those squared differences.

STEP 7

STEP 8

STEP 9

The **five-number summary** gives us a quick overview of the distribution of the fines.It consists of the minimum, Q1, median (Q2), Q3, and maximum.

STEP 10

First, we **arrange** the data in ascending order.The minimum is $15\$15$15 and the maximum is $250\$250$250.

STEP 11

The median is the middle value.Since we have 105\text{105}105 data points, the median is the 53rd53^{\text{rd}}53rd value, which is $30\$30$30.

STEP 12

Q1 is the median of the lower half of the data (excluding the median if the number of data points is odd).In our case, Q1 is the 27th27^{\text{th}}27th value, which is $30\$30$30.

STEP 13

Q3 is the median of the upper half of the data (excluding the median if the number of data points is odd).Q3 is the 80th80^{\text{th}}80th value, which is $40\$40$40.

STEP 14

So, our **five-number summary** is: Min=$15\text{Min} = \$15Min=$15, Q1=$30\text{Q1} = \$30Q1=$30, Median=$30\text{Median} = \$30Median=$30, Q3=$40\text{Q3} = \$40Q3=$40, Max=$250\text{Max} = \$250Max=$250.

STEP 15

A frequency distribution **organizes** the data into classes (or intervals) and shows how many data points fall into each class.We can create a table with the fine amounts and their corresponding frequencies.For example, the fine of $30\$30$30 appears 32\text{32}32 times.

STEP 16

A histogram is a visual representation of the frequency distribution.It uses bars to show the frequency of each class.The x-axis represents the fine amounts, and the y-axis represents the frequency.The height of each bar corresponds to the frequency of the corresponding fine amount.

STEP 17

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Start learning now

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1. Calculate the sample mean.
2. Calculate the sample standard deviation.
3. Find the five-number summary.
4. Create a frequency distribution.
5. Create a histogram.

To kick things off, let's find the sample mean, which is just the average of all the fines in our sample.
We add up all the fines and then divide by the number of fines, which is $\text{105}$ .

$\text{Sample Mean} = \frac{\text{Sum of fines}}{\text{Number of fines}} = \frac{3595}{105} \approx 34.24$ So, the sample mean is approximately $\$34.24$ .

Now, let's calculate the sample standard deviation, which tells us how spread out the fines are around the mean.
For each fine, we'll find the difference between the fine and the sample mean, square that difference, and add up all those squared differences.

The five-number summary gives us a quick overview of the distribution of the fines.
It consists of the minimum, Q1, median (Q2), Q3, and maximum.

First, we arrange the data in ascending order.
The minimum is $\$15$ and the maximum is $\$250$ .

The median is the middle value.
Since we have $\text{105}$ data points, the median is the $53^{\text{rd}}$ value, which is $\$30$ .

Q1 is the median of the lower half of the data (excluding the median if the number of data points is odd).
In our case, Q1 is the $27^{\text{th}}$ value, which is $\$30$ .

Q3 is the median of the upper half of the data (excluding the median if the number of data points is odd).
Q3 is the $80^{\text{th}}$ value, which is $\$40$ .

So, our five-number summary is: $\text{Min} = \$15$ , $\text{Q1} = \$30$ , $\text{Median} = \$30$ , $\text{Q3} = \$40$ , $\text{Max} = \$250$ .

A frequency distribution organizes the data into classes (or intervals) and shows how many data points fall into each class.
We can create a table with the fine amounts and their corresponding frequencies.
For example, the fine of $\$30$ appears $\text{32}$ times.

A histogram is a visual representation of the frequency distribution.
It uses bars to show the frequency of each class.
The x-axis represents the fine amounts, and the y-axis represents the frequency.
The height of each bar corresponds to the frequency of the corresponding fine amount.