Quantitative Reasoning Chapter 2 MTH105

  In chapter 2 we learn how descriptive statistics which is describing data through the method of graphical representation. Click for chapter 3.  The first example can be found below:

The following data represents the age of 30 lottery winners.

21	27	31	32	32	33
36	37	39	39	40	42
43	45	45	49	52	55
58	58	60	60	60	65
66	69	74	76	77	85

Complete the frequency distribution for the data.

Age	Frequency
20-29	Correct
30-39	Correct
40-49	Correct
50-59	Correct
60-69	Correct
70-79	Correct
80-89	Correct

For this question you need to count the number of frequency the age group appears in the top chart. 20-29 shows up twice. According to the chart 30-39 shows up 8 times. 40-49 shows up 6 times. 50-59 shows up 4 times. 60-69 shows up 6 times. 70-79 shows up 3 times. 80-89 shows up once.  Another example:

The following data represents the age of 30 lottery winners.

22	26	27	27	31	34
36	42	43	44	48	49
52	53	55	56	57	60
65	65	66	67	69	72
75	77	78	78	79	87

Complete the frequency distribution for the data.

Age	Frequency
20-29	4
30-39	3
40-49	5
50-59	5
60-69	6
70-79	6
80-89	1

Next Question:

A group of adults were asked how many children they have in their families. The bar graph below shows the number of adults who indicated each number of children.

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How many adults were questioned?

Correct

What percentage of the adults questioned had 1 children?

31.578947368421

Finding out how many adults were questioned you just count up the number. 

To figure out the percentage of adults that questioned one child you divide 6/19 then multiply to get the answer: 6/19= 0.3157894736842105 * 100 = 31.578947368421

Next Question:

The bar graph below shows the percentage of students who received each letter grade on their last English paper. The class contains 20 students.

[Graphs generated by this script: setBorder(28,45,0,10); initPicture(0,10,0,30);axes(1000,5,1,1000,5); fill="blue"; stroke="black";textabs([165,0],"Grades","above");rect([1,0],[3,30]);text([2,0],"A","below");rect([3,0],[5,20]);text([4,0],"B","below");rect([5,0],[7,30]);text([6,0],"C","below");rect([7,0],[9,20]);text([8,0],"D","below");]

What number of students earned a D on their paper?

Correct students

The bar graph represents the percentage of students who received each letter grade on their last English paper, and it shows the following distribution:

A: 30%

B: 20%

C: 30%

D: 20%

Now, let's calculate the number of students who received each grade based on the given percentages and the total number of students in the class (20 students):

Number of students who received an A:

(30% of 20) = 0.3 * 20 = 6 students

Number of students who received a B:

(20% of 20) = 0.2 * 20 = 4 students

Number of students who received a C:

(30% of 20) = 0.3 * 20 = 6 students

Number of students who received a D:

(20% of 20) = 0.2 * 20 = 4 students

So, based on the percentages and a class of 20 students:

6 students received an A.

4 students received a B.

6 students received a C.

4 students received a D.

Next Question:

Audrey categorized her spending for this month into four categories: Rent, Food, Fun, and Other. The percents she spent in each category are pictured here.

[Graphs generated by this script: initPicture(-5.5,5.5,-5.5,5.5);circle([0,0],5);line([0,0],[5,0]);line([0,0],[-.868,4.924]);text([1,2],'Rent','right');text([1,1],'30%','right');line([0,0],[-5,0]);text([-1,2],'Food','left');text([-1,1],'23%','left');line([0,0],[2.5,-4.33]);text([1,-1],'Fun','right');text([1,-2],'17%','right');text([-1,-1],'Other','left');text([-1,-2],'30%','left');]

If Audrey spent a total of $2600 this month, how much did she spend on Food?

$Correct

To find out how much Audrey spent on Food, you can calculate it based on the given percentage and the total spending of $2600.

Audrey spent 23% of her total budget on Food.

So, to calculate the amount spent on Food:

Amount Spent on Food = (23/100) * $2600

Now, calculate it:

Amount Spent on Food = (0.23) * $2600 = $598

Audrey spent $598 on Food this month.

Similar question:

Kara categorized her spending for this month into four categories: Rent, Food, Fun, and Other. The amounts she spent in each category are pictured here.

[Graphs generated by this script: initPicture(-5.5,5.5,-5.5,5.5);circle([0,0],5);line([0,0],[5,0]);line([0,0],[-.868,4.924]);text([1,2],'Rent','right');text([1,1],'$500','right');line([0,0],[-5,0]);text([-1,2],'Food','left');text([-1,1],'$400','left');line([0,0],[2.5,-4.33]);text([1,-1],'Fun','right');text([1,-2],'$300','right');text([-1,-1],'Other','left');text([-1,-2],'$600','left');]

How much total money did Kara spend this month? 1800

To find the total amount of money Kara spent this month, you need to add up the amounts she spent in each category:

Rent: $500

Food: $400

Fun: $300

Other: $600

Now, add these amounts together:

Total = $500 + $400 + $300 + $600 = $1800

Kara spent a total of $1800 this month.

Next Question:

Consider the data set

5

8

5

6

7

9

Find the average (mean): Correct

Find the median: Correct

To find the average (mean) and median for the given data set:

Data set: 5, 8, 5, 6, 7, 9

Average (Mean):

To find the average, sum up all the numbers and divide by the total count (number of data points).

Average = (5 + 8 + 5 + 6 + 7 + 9) / 6 = 40 / 6 = 6.67 (rounded to two decimal places)

So, the average (mean) is approximately 6.67.

Median:

To find the median, first, you need to arrange the data in ascending order:

5, 5, 6, 7, 8, 9

The median is the middle value in the sorted list. Since there are 6 data points (an even number), you take the average of the two middle values, which are 6 and 7.

Median = (6 + 7) / 2 = 13 / 2 = 6.5

So, the median is 6.5.

Next Question:

Two classes were given identical quizzes.

Class A had a mean score of 8 and a standard deviation of 0.3
Class B had a mean score of 8.3 and a standard deviation of 1.1

Which class scored better on average? Select an answer Class A Class B Correct

Which class had more consistent scores? Select an answer Class A Class B Correct

To determine which class scored better on average and which class had more consistent scores, you can compare the mean scores and standard deviations of the two classes.

Which class scored better on average?

Class B had a higher mean score of 8.3, while Class A had a mean score of 8.0. Therefore, Class B scored better on average.

Which class had more consistent scores?

To determine which class had more consistent scores, you can look at the standard deviation. A lower standard deviation indicates more consistent scores.

Class A had a standard deviation of 0.3, which is relatively low, suggesting that the scores in Class A were relatively consistent and close to the mean of 8.0.

Class B had a standard deviation of 1.1, which is higher than Class A's standard deviation. This higher standard deviation suggests that the scores in Class B were more spread out from the mean of 8.3, indicating less consistency.

So, Class A had more consistent scores as indicated by its lower standard deviation, while Class B scored better on average due to its higher mean score.

Next Question:

A sample was done, collecting the data below. Calculate the standard deviation, to one decimal place.

x
26
17
9
22
5

1.  1: Find the mean. 
2. To calculate the mean (average) of the data set:
3. 
   
4. Add up all the values:
5. 26 + 17 + 9 + 22 + 5 = 79
6. 
   
7. Divide the sum by the number of data points (which is 5 in this case):
8. Mean (x̄) = 79 / 5 = 15.8
9. 
   
10. So, the mean (average) of the data set is 15.8.
11. 
    
12.  2: Subtract the mean from each score.
14. To subtract 15.8 from each score in the given data set:
15. 
    
16. Subtract 15.8 from each value:
17. 
    
18. 26 - 15.8 = 10.2
19. 17 - 15.8 = 1.2
20. 9 - 15.8 = -6.8
21. 22 - 15.8 = 6.2
22. 5 - 15.8 = -10.8
23. 
    
24.  3: Square each deviation.
26. To square each of the adjusted values:
27. 
    
28. Square each value:
29. 
    
30. (10.2)^2 = 104.04
31. (1.2)^2 = 1.44
32. (-6.8)^2 = 46.24
33. (6.2)^2 = 38.44
34. (-10.8)^2 = 116.64
35. 
    
36.  4: Add the squared deviations.
38. To add the squared deviations:
39. 
    
40. Sum up the squared values:
41. 
    
42. 104.04 + 1.44 + 46.24 + 38.44 + 116.64 = 306.80
43. 
    
44. So, the sum of the squared deviations is 306.80.
45. 
    
46.  5: Divide the sum by one less than the number of data points.
48. To find the sample variance, you should divide the sum of the squared deviations by one less than the number of data points (which is 4 in this case since there are 5 data points). This is known as the Bessel correction for sample variance.
49. 
    
50. Sample Variance = Sum of Squared Deviations / (Number of Data Points - 1)
51. 
    
52. Sample Variance = 306.80 / (5 - 1) = 306.80 / 4 = 76.70
53. 
    
54. So, the sample variance is 76.70.
55. 
    
56.  6: Take the square root of the result from Step 5.
57. 
    

To find the sample standard deviation, you need to take the square root of the sample variance:

Sample Standard Deviation = √(Sample Variance)

Sample Standard Deviation = √(76.70) ≈ 8.76 (rounded to two decimal places)

So, the sample standard deviation is approximately 8.76 (rounded to two decimal places).

And rounding 8.76 to one decimal place we arrive at 8.8 our answer.  

Next question:

Find the mode for this list of numbers

74	81
24	58
34	81
50	41
3	73
77	59
22	98
76

Mode = 

To find the mode (the number that appears most frequently) in the given list of numbers, we'll need to count the frequency of each number:

74: 1 time

81: 2 times

24: 1 time

58: 1 time

34: 1 time

50: 1 time

41: 1 time

3: 1 time

73: 1 time

77: 1 time

59: 1 time

22: 1 time

98: 1 time

76: 1 time

The number 81 appears twice, which is more frequently than any other number in the list. Therefore, the mode is 81.

Next Question:

 Find the mean for this list of numbers

45	64
17	40
18	71
7	76
86	36
98	29
49	70
4

Mean = 

Next Question:

Find the median for this list of numbers

61	94
68	79
37	46
95	75
43	19
88	52
99	85
55

Median = 68

To find the median for the given list of numbers, you first need to arrange the numbers in ascending order and then determine the middle value.

Here's the list of numbers in ascending order:

19, 37, 43, 46, 52, 55, 61, 68, 75, 79, 85, 88, 94, 95, 99

Since there are 15 numbers in the list (an odd number), the median is the middle value, which in this case is 68.

Similar Question:

Calculate the average (mean) of the data shown, to two decimal places

x
8.6
21
26.5
17.7
9
28.8
23.4
20

Answer =19.38

Next Question:

Find the minimum for this list of numbers

86	21
65	69
53	29
22	31
39	57
58	49
43	11
20

Minimum = 11

To find the minimum value from each pair of numbers, you'll need to compare the values within each pair and select the smaller value. Then, find the minimum among these smaller values. Here's the comparison for each pair:

Pair 1: (86, 21)

Minimum of Pair 1 = 21

Pair 2: (65, 69)

Minimum of Pair 2 = 65

Pair 3: (53, 29)

Minimum of Pair 3 = 29

Pair 4: (22, 31)

Minimum of Pair 4 = 22

Pair 5: (39, 57)

Minimum of Pair 5 = 39

Pair 6: (58, 49)

Minimum of Pair 6 = 49

Pair 7: (43, 11)

Minimum of Pair 7 = 11

Pair 8: (20)

Since this pair has only one number, the minimum is 20.

Now, find the minimum among these minimum values:

Minimum = min(21, 65, 29, 22, 39, 49, 11, 20) = 11

So, the minimum value in the given pairs of numbers is 11.

Next Question:

Find the first quartile for this list of numbers

100	40
4	67
17	72
20	27
68	81
24	94
49	9
39

Quartile 1 = 20

To find the first quartile (Q1) for this list of numbers, you first need to calculate the median of the lower half of the data. Here's the procedure:

Arrange the data in ascending order:

4, 9, 17, 20, 24, 27, 39, 40, 49, 68, 72, 81, 94, 100

Since there are 14 data points, calculate the median for the lower half of the data, which includes the first 7 values:

Median of Lower Half = (17 + 20 + 24 + 27 + 39 + 4 + 9) / 7 = 140 / 7 = 20

So, the first quartile (Q1) is 20.

Similar question:

Find the first quartile for this list of numbers

10	19
62	42
80	52
37	81
35	2
98	65
40	32
27

Quartile 1 = 27

Next Question:

Find the third quartile for this list of numbers

58	28
67	54
86	19
48	70
93	15
100	4
36	69
12

Quartile 3 = 70

First, arrange the data in ascending order:

4, 12, 15, 19, 28, 36, 48, 54, 58, 67, 69, 70, 86, 93, 100

Since there are 15 data points, we need to calculate the median of the upper half of the data, which includes the last 7 values:

Median of Upper Half = (58 + 67 + 69 + 70 + 86 + 93 + 100) / 7 = 563 / 7 = 80.43 (rounded to two decimal places)

The third quartile (Q3) is the value that divides the upper 25% of the data. In this case, since there are 15 data points, the upper 25% corresponds to the last 25% of the data. Therefore, Q3 is the median of the upper half, which is 70.

Next Question

Find the maximum for this list of numbers

88	6
15	38
49	81
33	29
53	58
30	9
42	62
27

Maximum = 88

To find the maximum value from each pair of numbers, you'll need to compare the values within each pair and select the larger value. Then, find the maximum among these larger values. Here's the comparison for each pair:

Pair 1: (88, 6)

Maximum of Pair 1 = 88

Pair 2: (15, 38)

Maximum of Pair 2 = 38

Pair 3: (49, 81)

Maximum of Pair 3 = 81

Pair 4: (33, 29)

Maximum of Pair 4 = 33

Pair 5: (53, 58)

Maximum of Pair 5 = 58

Pair 6: (30, 9)

Maximum of Pair 6 = 30

Pair 7: (42, 62)

Maximum of Pair 7 = 62

Pair 8: (27)

Since this pair has only one number in it, the maximum is 27.

Now, find the maximum among these maximum values:

Maximum = max(88, 38, 81, 33, 58, 30, 62, 27) = 88

Next Question:

Find the range for this list of numbers

77	91
14	71
35	86
24	31
83	41
30	95
82	92
44

Range = 81

The range of a set of numbers is the difference between the maximum and minimum values. In this case, you've already identified the maximum and minimum values from each pair of numbers.

Maximum values:

91, 71, 86, 31, 83, 95, 92, 44

Minimum values:

77, 14, 35, 24, 41, 30, 82, 44

Now, find the maximum and minimum values from the maximum and minimum lists:

Maximum value (from maximum list): 95

Minimum value (from minimum list): 14

Now, calculate the range:

Range = Maximum Value - Minimum Value = 95 - 14 = 81

Next Question:

The line graph below shows the number of times Maria drove her car each day of the week.

[Graphs generated by this script: initPicture(-1,8,-2,10);axes(1,1,1,1,1);marker="dot";line([1,2],[2,8]);line([2,8],[3,5]);line([3,5],[4,9]);line([4,9],[5,4]);line([5,4],[6,1]);line([6,1],[7,7]);text([4,-1],"Day of Week","below");]

How many times did Maria drive her car on day 4? 9

You just plot the day of the week which is the bottom line with the dot which lands on the top line of 9

Next Question:

Kara categorized her spending for this month into four categories: Rent, Food, Fun, and Other. The amounts she spent in each category are pictured here.

[Graphs generated by this script: initPicture(-5.5,5.5,-5.5,5.5);circle([0,0],5);line([0,0],[5,0]);line([0,0],[-.868,4.924]);text([1,2],'Rent','right');text([1,1],'$472','right');line([0,0],[-5,0]);text([-1,2],'Food','left');text([-1,1],'$378','left');line([0,0],[2.5,-4.33]);text([1,-1],'Fun','right');text([1,-2],'$283','right');text([-1,-1],'Other','left');text([-1,-2],'$567','left');]

What percent of her total spending did she spend on Fun? Answer to the nearest whole percent. 17

To find the percentage of her total spending that she spent on Fun, you can use the following calculation:

Percentage spent on Fun = (Fun / Total spending) * 100

Fun: $283

Total spending: $472 (Rent) + $378 (Food) + $283 (Fun) + $567 (Other) = $1700

Percentage spent on Fun = ($283 / $1700) * 100 ≈ 16.65%

So, she spent approximately 16.65% of her total spending on Fun, rounded to the nearest whole percent.

And we round 16.65% to 17 to make it a whole number. 

Similar questions:

Kara categorized her spending for this month into four categories: Rent, Food, Fun, and Other. The amounts she spent in each category are pictured here.

[Graphs generated by this script: initPicture(-5.5,5.5,-5.5,5.5);circle([0,0],5);line([0,0],[5,0]);line([0,0],[-.868,4.924]);text([1,2],'Rent','right');text([1,1],'$306','right');line([0,0],[-5,0]);text([-1,2],'Food','left');text([-1,1],'$244','left');line([0,0],[2.5,-4.33]);text([1,-1],'Fun','right');text([1,-2],'$183','right');text([-1,-1],'Other','left');text([-1,-2],'$367','left');]

What percent of her total spending did she spend on Food? Answer to the nearest whole percent. 22

Kara categorized her spending for this month into four categories: Rent, Food, Fun, and Other. The amounts she spent in each category are pictured here.

[Graphs generated by this script: initPicture(-5.5,5.5,-5.5,5.5);circle([0,0],5);line([0,0],[5,0]);line([0,0],[-.868,4.924]);text([1,2],'Rent','right');text([1,1],'$278','right');line([0,0],[-5,0]);text([-1,2],'Food','left');text([-1,1],'$222','left');line([0,0],[2.5,-4.33]);text([1,-1],'Fun','right');text([1,-2],'$167','right');text([-1,-1],'Other','left');text([-1,-2],'$333','left');]

What percent of her total spending did she spend on Fun? Answer to the nearest whole percent. 17

Next question:

A political scientist surveys 26 of the current 130 representatives in a state's legislature.

What is the size of the sample: 

What is the size of the population: 

Next Question:

In a survey of 1462 adults, 71% responded "yes" to the survey question.

How many adults answered "yes"? (round to the nearest whole person as needed) 1,038

To find out how many adults answered "yes" in the survey, you can calculate 71% of the total number of adults surveyed.

71% of 1462 = (71/100) * 1462 ≈ 1038.02

Rounded to the nearest whole person, approximately 1,038 adults answered "yes" to the survey question.

Next Question:

Convert the fraction 7/10 to an equivalent percentage.

Report exact answer (give all decimal places without rounding).

${\displaystyle \frac{\mathrm{<br>}}{10}=}$ Correct %

7/10= 0.7 * 100 = 70

Next Question:

Convert the fraction 113/8052 to an equivalent percentage.

Report your answer accurate to one decimal place.

${\displaystyle \frac{113}{8052}=}$ Correct

113/8052 = 0.0140337804272231 * 100 = 1.403378042722305

Next Question:

Convert 0.1% to an equivalent decimal.

${\displaystyle 0.1}$0.1% = 0.01

To convert 0.1% to an equivalent decimal, you move the decimal point two places to the left because 1% is equivalent to 0.01 as a decimal.

So, 0.1% is equivalent to 0.001 as a decimal.

Next Question:

Convert 1.2% to an equivalent decimal.

To convert 1.2% to an equivalent decimal, you move the decimal point two places to the left because 1% is equivalent to 0.01 as a decimal.

So, 1.2% is equivalent to 0.012 as a decimal.

Next Question:

What is ${\displaystyle 76\%}$76% of 850?

Ans = 646

To find 76% of 850, you can calculate it as follows:

(76/100) * 850 = 0.76 * 850 = 646

So, 76% of 850 is equal to 646.

Next Question:

Convert 0.047 to an equivalent percentage.

To convert the decimal 0.047 to an equivalent percentage, you multiply it by 100:

0.047 * 100 = 4.7%

So, 0.047 is equivalent to 4.7% as a percentage.

Next Question:

Ages	Number of students
15-18	2
19-22	4
23-26	8
27-30	9
31-34	6
35-38	3

Based on the frequency distribution above, is 22.5 a:

• Class boundary
• Lower class limit
• Upper class limit
• Class width
• Class midpoint

${\displaystyle 1.2\%=}$ 

In the given frequency distribution:

Ages | Number of students

15-18 | 2

19-22 | 4

23-26 | 8

27-30 | 9

31-34 | 6

35-38 | 3

Class boundary:

A class boundary separates one class from another. For example, between 18 and 19, there is a class boundary, and 22.5 falls into this boundary.

Lower class limit:

The lower class limit is the smallest value in each class interval. For example, in the "19-22" class, the lower class limit is 19.

Upper class limit:

The upper class limit is the largest value in each class interval. For example, in the "19-22" class, the upper class limit is 22.

Class width:

The class width is the range covered by each class interval. It is the difference between the upper class limit and the lower class limit. For example, in the "19-22" class, the class width is 22 - 19 = 3.

Class midpoint:

The class midpoint is the middle value of each class interval. To find it, you can take the average of the lower class limit and the upper class limit. For example, in the "19-22" class, the class midpoint would be (19 + 22) / 2 = 20.5.

So, 22.5 is indeed a class boundary in the distribution, particularly between the "19-22" and "23-26" classes. 

Similar questions:

Ages	Number of students
15-18	2
19-22	4
23-26	8
27-30	9
31-34	6
35-38	3

Based on the frequency distribution above, is 19 a:

• Class width
• Class midpoint
• Lower class limit
• Class boundary
• Upper class limit

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

Lengths (mm)	Frequency
180 - 194	1
195 - 209	16
210 - 224	71
225 - 239	108
240 - 254	83
255 - 269	18
270 - 284	3

What is the upper class limit for the fourth class?

upper class limit =239

Next Question:

Ages	Number of students
15-18	6
19-22	5
23-26	4
27-30	9
31-34	6
35-38	2

Based on the frequency distribution above, find the relative frequency for the class with lower class limit 19

Relative Frequency =15.625

To find the relative frequency for the class with a lower class limit of 19, you need to divide the number of students in that class by the total number of students in all classes.

In this case, the class with a lower class limit of 19 is "19-22," and it has 5 students.

The total number of students is the sum of students in all classes:

6 (15-18) + 5 (19-22) + 4 (23-26) + 9 (27-30) + 6 (31-34) + 2 (35-38) = 32

Now, calculate the relative frequency:

Relative Frequency = (Number of students in the class / Total number of students) * 100

Relative Frequency = (5 / 32) * 100 ≈ 15.625%

So, the relative frequency for the class with a lower class limit of 19 is approximately 15.625%.

Similar Question:

Ages	Number of students
15-18	8
19-22	6
23-26	2
27-30	3
31-34	9
35-38	8

Based on the frequency distribution above, find the relative frequency for the class with lower class limit 19

Relative Frequency =16.666666666667

Next Question:

Ages	Number of students
15-18	7
19-22	9
23-26	8
27-30	2
31-34	4
35-38	6

Based on the frequency distribution above, find the cumulative frequency for the class with lower class limit 27

Cumulative Frequency =26

To find the cumulative frequency for the class with a lower class limit of 27, you'll need to add up the frequencies of all the classes with lower class limits less than or equal to 27.

In this case, the class with a lower class limit of 27 is "27-30," and it has a frequency of 2.

Now, let's calculate the cumulative frequency:

Cumulative Frequency = Sum of frequencies of all classes with lower class limits less than or equal to 27

Cumulative Frequency = 7 (15-18) + 9 (19-22) + 8 (23-26) + 2 (27-30)

Cumulative Frequency = 7 + 9 + 8 + 2 = 26

So, the cumulative frequency for the class with a lower class limit of 27 is 26.

Data was collected for 259 randomly selected 10 minute intervals. For each ten-minute interval, the number of people entering the atrium of a large mall were recorded. The data is summarized in the table below.

Number   of Guests	Frequency
120 – 124	81
125 – 129	95
130 – 134	14
135 – 139	22
140 – 144	47

What is the class width for this GFDT?

Class width =5

The class width for a grouped frequency distribution table (GFDT) is the difference between the upper class limit and the lower class limit of each class interval.

In your provided data:

For the class "120 - 124," the lower class limit is 120, and the upper class limit is 124.

For the class "125 - 129," the lower class limit is 125, and the upper class limit is 129.

For the class "130 - 134," the lower class limit is 130, and the upper class limit is 134.

For the class "135 - 139," the lower class limit is 135, and the upper class limit is 139.

For the class "140 - 144," the lower class limit is 140, and the upper class limit is 144.

To find the class width, you can subtract the lower class limit from the upper class limit for any of the classes:

Class width = Upper Class Limit - Lower Class Limit

Let's use the first class as an example:

Class width = 124 (Upper Class Limit) - 120 (Lower Class Limit) = 5

Similar question:

Data was collected for 278 randomly selected 10 minute intervals. For each ten-minute interval, the number of people entering the atrium of a large mall were recorded. The data is summarized in the table below.

Number   of Guests	Frequency
80 – 99	77
100 – 119	29
120 – 139	96
140 – 159	65
160 – 179	11

What is the class width for this GFDT?

Class width =20

Next Question:

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

Lengths (mm)	Frequency
160 - 162	1
163 - 165	16
166 - 168	71
169 - 171	108
172 - 174	83
175 - 177	18
178 - 180	3

What is the class midpoint for the seventh class?

class midpoint = 179

Class midpoint = (Lower Class Limit + Upper Class Limit) / 2

Class midpoint = (175 + 177) / 2 = 352 / 2 = 176

Next Question:

120 psychology students took a standardized test. The scores are summarized in the GFDT below.

Scores	Frequency
120 - 124	14
125 - 129	20
130 - 134	17
135 - 139	12
140 - 144	12
145 - 149	45

The scores are also described in the cumulative table shown below.

Scores	Frequency
less than 125	14
less than 130	34
less than 135	51
less than 140
less than 145	75
less than 150	120

What is the missing value in the cumulative GFDT?

answer =63

Cumulative Frequency (less than 140) = Cumulative Frequency (less than 135) - Cumulative Frequency (less than 139) = 51 - 12 = 39

Next Question:

n psychology students took a standardized test. The scores are summarized in the GFDT below.

Scores	Frequency
40 - 59	14
60 - 79	17
80 - 99	18
100 - 119
120 - 139	19
140 - 159	39

The scores are also described in the cumulative table shown below.

Scores	Frequency
less than 60	14
less than 80	31
less than 100	49
less than 120	62
less than 140	81
less than 160	${\displaystyle �}$

What is the missing value in the GFDT?

answer = 13

To find the missing value in the grouped frequency distribution table (GFDT), we can look at the cumulative table and see that it provides cumulative frequencies for various score ranges. We need to determine the cumulative frequency for "less than 100" to find the missing value in the GFDT.

Cumulative Frequency (less than 100) = 49

Now, to find the missing value in the GFDT, we can calculate the frequency for the "100 - 119" score range:

Frequency (100 - 119) = Cumulative Frequency (less than 100) - Cumulative Frequency (less than 120)

Frequency (100 - 119) = 62 - 49  = 13

Next Question:

50 psychology students took a standardized test. The scores are summarized in the GFDT below.

Scores	Frequency
40 - 49	11
50 - 59	9
60 - 69	9
70 - 79	8
80 - 89	13

You would like to build a relative frequency distribution for this data set to easily compare it to another class of students.

What is the relative frequency for the second class, 50 - 59?

Enter answer as a percent (but do not include the percent symbol).

relative frequency = 18

To calculate the relative frequency for the class "50 - 59," you need to divide the frequency of that class by the total number of students and then express the result as a percentage.

The frequency for the class "50 - 59" is 9.

The total number of students is the sum of all the frequencies:

11 (40 - 49) + 9 (50 - 59) + 9 (60 - 69) + 8 (70 - 79) + 13 (80 - 89) = 50

Now, calculate the relative frequency:

Relative Frequency (50 - 59) = (Frequency for 50 - 59 / Total number of students) * 100

Relative Frequency (50 - 59) = (9 / 50) * 100 = 18%

So, the relative frequency for the class "50 - 59" is 18%.

Next Question:

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

Lengths (mm)	Frequency
40 - 54	1
55 - 69	16
70 - 84	71
85 - 99	108
100 - 114	83
115 - 129	18
130 - 144	3

What is the lower class limit for the fourth class?

lower class limit =85

The lower class limit for the fourth class, which is "85 - 99," is the smallest value within that class interval.

In this case, the lower class limit is 85

Next Question:

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

Lengths (mm)	Frequency
200 - 214	1
215 - 229	16
230 - 244	71
245 - 259	108
260 - 274	83
275 - 289	18
290 - 304	3

What is the upper class limit for the second class?

upper class limit =229

The upper class limit for the second class, which is "215 - 229," is the largest value within that class interval.

In this case, the upper class limit is 229

Similar Question:

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

Lengths (mm)	Frequency
200 - 202	1
203 - 205	16
206 - 208	71
209 - 211	108
212 - 214	83
215 - 217	18
218 - 220	3

What is the lower class limit for the fourth class?

lower class limit = 209

Next Question: 

Determine the distribution of the data pictured below

[Graphs generated by this script: setBorder(61,40,20,5); initPicture(0.1,8,0,4.8);axes(1000,1,1,1000,1); fill="blue"; stroke="black"; textabs([165,0],"data","above");line([1,-0.096],[1,0.096]); text([1,0],"1","below");line([2,-0.096],[2,0.096]); text([2,0],"2","below");line([3,-0.096],[3,0.096]); text([3,0],"3","below");line([4,-0.096],[4,0.096]); text([4,0],"4","below");line([5,-0.096],[5,0.096]); text([5,0],"5","below");line([6,-0.096],[6,0.096]); text([6,0],"6","below");line([7,-0.096],[7,0.096]); text([7,0],"7","below");line([8,-0.096],[8,0.096]); text([8,0],"8","below");textabs([0,115],"Frequency","right",90);rect([1,0],[2,0.7]);rect([2,0],[3,2.01]);rect([3,0],[4,3.95]);rect([4,0],[5,4.8]);rect([5,0],[6,4.15]);rect([6,0],[7,1.77]);rect([7,0],[8,1.17]);]

• Uniform
• Bell-shaped
• Skewed-right
• Skewed-left

In the provided, the data is symmetrically distributed around the center (values 3 and 4), resembling a bell-shaped curve.

Next Question:

The line graph below shows the number of times Maria drove her car each day of the week.

[Graphs generated by this script: initPicture(-1,8,-2,10);axes(1,1,1,1,1);marker="dot";line([1,7],[2,4]);line([2,4],[3,1]);line([3,1],[4,9]);line([4,9],[5,2]);line([5,2],[6,6]);line([6,6],[7,5]);text([4,-1],"Day of Week","below");]

How many times did Maria drive her car on day 4? 9

Next Question:

Determine the distribution of the data pictured below

[Graphs generated by this script: setBorder(54,40,20,5); initPicture(0,7.5,0,30);axes(1000,5,1,1000,5); fill="blue"; stroke="black"; textabs([165,0],"data","above");line([0.5,-0.6],[0.5,0.6]); text([0.5,0],"0.5","below");line([1.5,-0.6],[1.5,0.6]); text([1.5,0],"1.5","below");line([2.5,-0.6],[2.5,0.6]); text([2.5,0],"2.5","below");line([3.5,-0.6],[3.5,0.6]); text([3.5,0],"3.5","below");line([4.5,-0.6],[4.5,0.6]); text([4.5,0],"4.5","below");line([5.5,-0.6],[5.5,0.6]); text([5.5,0],"5.5","below");line([6.5,-0.6],[6.5,0.6]); text([6.5,0],"6.5","below");line([7.5,-0.6],[7.5,0.6]); text([7.5,0],"7.5","below");textabs([0,115],"Frequency","right",90);rect([0.5,0],[1.5,27]);rect([1.5,0],[2.5,30]);rect([2.5,0],[3.5,22]);rect([3.5,0],[4.5,9]);rect([4.5,0],[5.5,6]);rect([5.5,0],[6.5,0]);rect([6.5,0],[7.5,1]);]

• Uniform
• Bell-shaped
• Skewed-right
• Skewed-left

We see the bars lower towards the right which means its Skewed-right.

Next Question:

Data was collected for 189 randomly selected 10 minute intervals. For each ten-minute interval, the number of people entering the atrium of a large mall were recorded. The data is summarized in the histogram below.

[Graphs generated by this script: setBorder(54,40,20,5); initPicture(95.5,125,0,76);axes(1000,20,1,1000,20); fill="blue"; stroke="black"; textabs([165,0],"Number of People in 10 min","above");line([100,-1.52],[100,1.52]); text([100,0],"100","below");line([105,-1.52],[105,1.52]); text([105,0],"105","below");line([110,-1.52],[110,1.52]); text([110,0],"110","below");line([115,-1.52],[115,1.52]); text([115,0],"115","below");line([120,-1.52],[120,1.52]); text([120,0],"120","below");line([125,-1.52],[125,1.52]); text([125,0],"125","below");textabs([0,115],"Frequency","right",90);rect([100,0],[105,64]);rect([105,0],[110,29]);rect([110,0],[115,6]);rect([115,0],[120,14]);rect([120,0],[125,76]);]

What is the class width for this histogram (and corresponding GFDT)? Note: Each class contains its lower class limit, but not its upper class limit.

Class width = 5

To find the class width for this histogram, you can look at the gaps between the lower class limits of adjacent classes. In this histogram, the lower class limits are given for each class, but the upper class limits are not explicitly provided.

Let's examine the gaps:

From 100 to 105, the gap is 105 - 100 = 5.

From 105 to 110, the gap is 110 - 105 = 5.

From 110 to 115, the gap is 115 - 110 = 5.

From 115 to 120, the gap is 120 - 115 = 5.

From 120 to 125, the gap is 125 - 120 = 5.

The gaps between the lower class limits of all adjacent classes are consistent and equal to 5.

So, the class width for this histogram (and corresponding grouped frequency distribution table) is 5.

Next Question:

Calculate the median of the data shown, to two decimal places

x
14.9
10.4
28.2
10.5
13.7

To calculate the median of the given data, you'll first need to arrange the data in ascending order. Once the data is sorted, you can find the middle value as the median.

Here's the data in ascending order:

10.4, 10.5, 13.7, 14.9, 28.2

Since there are five data points, the median will be the middle value. In this case, the median is 13.7.

So, the median of the data is 13.7 (rounded to two decimal places).

Next Question:

In a city with three high schools, all the ninth graders took a Standardized Test, with these results:

High School	Mean score on test	Number of ninth graders
Glenwood	84	284
Central City	96	330
Lincoln High	69	158

The city's PR manager, who never took statistics, claimed the mean score of all ninth graders in the city was 83 . Of course, that is incorrect. What is the mean score for all ninth graders in the city? Round to one decimal place.

mean of all ninth grader's scores =86.1

To calculate the mean score for all ninth graders in the city, you can use the weighted average formula, where you multiply each high school's mean score by the number of ninth graders from that high school, sum these products, and then divide by the total number of ninth graders in the city.

Mean score for Glenwood: 84

Number of ninth graders at Glenwood: 284

Mean score for Central City: 96

Number of ninth graders at Central City: 330

Mean score for Lincoln High: 69

Number of ninth graders at Lincoln High: 158

Now, calculate the weighted average:

Weighted Average = [(Mean score at Glenwood * Number of ninth graders at Glenwood) + (Mean score at Central City * Number of ninth graders at Central City) + (Mean score at Lincoln High * Number of ninth graders at Lincoln High)] / Total number of ninth graders

Weighted Average = [(84 * 284) + (96 * 330) + (69 * 158)] / (284 + 330 + 158)

Weighted Average = (23856 + 31680 + 10962) / 772

Weighted Average = 66598 / 772

Weighted Average ≈ 86.13 (rounded to one decimal place)

So, the mean score for all ninth graders in the city is approximately 86.1 when rounded to one decimal place, not 83 as claimed by the PR manager.

Next Question:

On January 1, 2009, Rachael owed $13,851 to her friend Salomé, who was kind enough not to charge Rachael any interest. Each month during 2009, Rachael paid Salomé some of the money she owed. If Rachael still owed Salomé $9,807 on January 1, 2010, what was the average amount of Rachael's monthly payments?

$337

To find the average amount of Rachael's monthly payments, you can calculate the total amount she paid off over the course of the year (from January 1, 2009, to January 1, 2010) and then divide it by the number of months (12).

Initial amount owed on January 1, 2009: $13,851

Amount remaining on January 1, 2010: $9,807

Total amount paid off: $13,851 - $9,807 = $4,044

Now, divide the total amount paid off by the number of months (12) to find the average monthly payment:

Average Monthly Payment = Total amount paid off / Number of months

Average Monthly Payment = $4,044 / 12 = $337

So, the average amount of Rachael's monthly payments was $337.

Next Question:

In a mid-size company, the distribution of the number of phone calls answered each day by each of the 12 receptionists is bell-shaped and has a mean of 62 and a standard deviation of 8. Using the empirical rule (as presented in the book), what is the approximate percentage of daily phone calls numbering between 54 and 70?

Do not enter the percent symbol.
ans =68

To find the approximate percentage of daily phone calls numbering between 54 and 70 using the empirical rule (also known as the 68-95-99.7 rule), you can consider that:

About 68% of the data falls within one standard deviation of the mean.

About 95% of the data falls within two standard deviations of the mean.

About 99.7% of the data falls within three standard deviations of the mean.

In your case, the mean is 62, and the standard deviation is 8.

First, let's find one standard deviation above and below the mean:

One Standard Deviation Below the Mean: 62 - 8 = 54

One Standard Deviation Above the Mean: 62 + 8 = 70

Now, you want to find the percentage of daily phone calls between 54 and 70, which is within one standard deviation above and below the mean. This corresponds to the central 68% of the data.

So, the approximate percentage of daily phone calls numbering between 54 and 70 is 68%.

Next Question:

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 48 and a standard deviation of 8. Using the empirical rule (as presented in the book), what is the approximate percentage of lightbulb replacement requests numbering between 32 and 48?

Do not enter the percent symbol.
ans = 47.5

Next Question:

Adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the z-score of a man 76.9 inches tall. (to 2 decimal places)

The z-score is a measure of how many standard deviations an individual's data point (in this case, height) is away from the mean (average) of the dataset.

Z = x- y / o

Z is the z-score.

X is the individual's height.

y is the mean height.

o is the standard deviation.

Z = 76.9 -69.0/2.8

Z ≈ 7.9/2.9

Z ≈ 2.82

Here's how you calculate the z-score step by step:

Start with the individual's height: 76.9 inches.

Subtract the mean height of adult men: 69.0 inches.

Divide the result by the standard deviation: 2.8 inches.

So, the calculation is:

Z = 76.9-69.0/2.8 ≈ 7.9/2.9 ≈2.82

Next Question:

On a recent quiz, the class mean was 73 with a standard deviation of 3.1. Calculate the z-score (to 2 decimal places) for a person who received score of 75.

z-score: Correct
Is this unusual?

• Not Unusual
• Unusual

The z-score for a person who received a score of 75 in the class with a mean of 73 and a standard deviation of 3.1 is indeed 0.65 (rounded to two decimal places), as you correctly calculated.

Now, let's determine if this z-score is considered unusual. In general, z-scores can help us assess how far an individual's score is from the mean in terms of standard deviations. Here's a rough guideline:

A z-score of 0 means the score is exactly at the mean.

Positive z-scores indicate scores above the mean.

Negative z-scores indicate scores below the mean.

In this case, a z-score of 0.65 means the person's score is approximately 0.65 standard deviations above the mean. This is not very unusual, as it's less than 1 standard deviation away from the mean. In many cases, z-scores within the range of -1 to 1 are considered relatively typical or not unusual.

So, a z-score of 0.65 is not unusual.

Next Question: 

Convert the fraction 7/25 to an equivalent percentage.

Report exact answer (give all decimal places without rounding).

Ans = 28

Next Question:

Convert 2.8% to an equivalent decimal.

Ans= 0.028

Next Question:

Consider the data set

7

3

1

7

4

3

1

Find the average (mean): Correct

Round 3 places after the decimal if needed.

Find the median: 3

Next Question:

In a mid-size company, the distribution of the number of phone calls answered each day by each of the 12 receptionists is bell-shaped and has a mean of 35 and a standard deviation of 9. Using the empirical rule (as presented in the book), what is the approximate percentage of daily phone calls numbering between 26 and 44?

Do not enter the percent symbol.
ans =68

To find the approximate percentage of daily phone calls numbering between 26 and 44 using the empirical rule (also known as the 68-95-99.7 rule), you can consider that:

About 68% of the data falls within one standard deviation of the mean.

About 95% of the data falls within two standard deviations of the mean.

About 99.7% of the data falls within three standard deviations of the mean.

In your case, the mean is 35, and the standard deviation is 9.

First, let's find one standard deviation below and above the mean:

One Standard Deviation Below the Mean: 35 - 9 = 26

One Standard Deviation Above the Mean: 35 + 9 = 44

So, the approximate percentage of daily phone calls numbering between 26 and 44 is within one standard deviation from the mean, which corresponds to the central 68% of the data.

The answer is 68%.

Project #2 - M&M Proportions B

Must show calculations/formula for credit. Any answer given with no calculations shown will result in no credit for that answer.

 

1.       Suppose you open a bag of M&Ms and count the number of M&Ms you have of each color. Suppose there were the following:

Qty 11 blue

Qty 6 orange

Qty 4 yellow

Qty 5 green

Qty 3 brown

Qty 5 red

 

2.       Calculate the proportion of each color (round to 3 decimals) please show calculation and answer in proportion cell of table example 21/57 = .368

 

Color	Proportion
Blue	11/34 = 0.324
Orange	6/34 = 0.176
Yellow	4/34 = 0.118
Green	5/34 = 0.147
Brown	3/34 = 0.088
Red	5/34 = 0.147

 

 

3.       Suppose the entire class put all the M&Ms together on the table and following occurred:

 

Qty 45 blue

Qty 47 orange

Qty 31 yellow

Qty 28 green

Qty 12 brown

Qty 19 red

4.       For the entire class, calculate the proportion of each color (round to 3 decimals) please show calculation and answer in proportion cell of table example 21/57 = .368

 

Color	Proportion
Blue	45/182 = 0.247
Orange	47/182 =0.258
Yellow	31/182 =0.170
Green	28/182 =0.154
Brown	12/182 =0.066
Red	19/182 =0.104

 

 

5.       What conclusions can you draw from the data gathered from one small bag compared to several small bags of M&Ms? That there’s never an equal number of each color of M&M they are always slightly not proportional.

6.       A 5-pound bulk bag of M&Ms holds approximately 2775 pieces. How many of each color should we expect to find in the bag based on the results of the proportions from the entire class? (approximate number of each color to be rounded to the nearest whole M&M) please show calculation and answer in Approximate number of M&Ms cell of table example 2900*0.368 = 1067

 

 

 

Color	Approximate number of M&Ms
Blue	2775*0.247 =685
Orange	2775*0.258 =716
Yellow	2775*0.170 =472
Green	2775*0.154 =427
Brown	2775*0.066 =183
Red	2775*0.104 =289

 

 

 

7.       A bag of M&Ms is dumped on the table and you count 86 blue pieces.  Approximately how many M&Ms do you expect to be in the entire bag based on the proportions of the entire class?  Round to nearest whole M&M

Show calculation and answer

 Blue 0.247*86 = 21

Orange 0.258*86 = 22

Yellow 0.170*86 =15

Green 0.154*86 =13

Brown 0.066*86 =6

Red 0.104*86 =9

 8.       According to Mars, the makers of M&Ms, their factory produces 25% blue, 25% orange, and 12.5% for each of the remaining colors. How do the observed values based on the entire class compare to these? It seems like yellow isn’t that far off from Orange and blue but the percentage amounts are pretty close to the way it is.

9.       What are some of the reasons why our bags would differ from the expected distribution? A machine bags them so there isn’t an exact counter that puts each color in they are likely distributed in a random fashion to equal a total weight count.

10.   Why would Mars not put equal amounts of each color in every bag? It’s done by a machine likely and they aren’t counting out each color.