Measures of Central Tendency (Mean, Median, Mode)

37 soru

Soru 1Soru

A research study records five numerical observations: 1212, 1616, 2020, 2020, and xx, where xx is a positive integer. If the arithmetic mean of these five observations is equal to their median, which of the following values could be xx? Select all such values.

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Cevap: 1212; 1717; 3232

Cevap

The possible values of xx are 1212, 1717, and 3232.
The mean of the dataset is given by 68+x5\frac{68+x}{5}. Depending on where xx falls relative to the sorted known values (12,16,20,2012, 16, 20, 20), the median can be 1616 (if x16x \le 16), xx (if 16x2016 \le x \le 20), or 2020 (if x20x \ge 20). Equating the mean to the median in each case yields three valid solutions: x=12x = 12, x=17x = 17, and x=32x = 32.

Adım Adım Çözüm

1
Express the arithmetic mean in terms of xx.
The sum of the five numbers is 12+16+20+20+x=68+x12 + 16 + 20 + 20 + x = 68 + x. Therefore, the mean is 68+x5\frac{68 + x}{5}.
The mean of nn numbers is the total sum divided by nn.
2
Analyze Case 1 where x16x \le 16.
The sorted dataset is x,12,16,20,20x, 12, 16, 20, 20 (or 12,x,16,20,2012, x, 16, 20, 20). The median (3rd value) is 1616. Setting mean = median gives 68+x5=16    68+x=80    x=12\frac{68 + x}{5} = 16 \implies 68 + x = 80 \implies x = 12. Since 121612 \le 16, x=12x = 12 is valid.
The median of a 5-element set is the 3rd element when ordered.
3
Analyze Case 2 where 16x2016 \le x \le 20.
The sorted dataset is 12,16,x,20,2012, 16, x, 20, 20. The median is xx. Setting mean = median gives 68+x5=x    68+x=5x    4x=68    x=17\frac{68 + x}{5} = x \implies 68 + x = 5x \implies 4x = 68 \implies x = 17. Since 16172016 \le 17 \le 20, x=17x = 17 is valid.
When xx lies between 16 and 20, xx itself becomes the 3rd element of the ordered set.
4
Analyze Case 3 where x20x \ge 20.
The sorted dataset is 12,16,20,20,x12, 16, 20, 20, x. The median is 2020. Setting mean = median gives 68+x5=20    68+x=100    x=32\frac{68 + x}{5} = 20 \implies 68 + x = 100 \implies x = 32. Since 322032 \ge 20, x=32x = 32 is valid.
When x20x \ge 20, 20 is the 3rd element of the ordered set.

Anahtar Kavram

Solving for missing observations where mean equals median requires analyzing how the position of the variable affects the sorted order and the resulting median value.
Soru 2Soru

A quality control department recorded the processing time, in days, for a batch of requests. The table below displays the frequency distribution of the processing times:

Processing Time (Days)Number of Requests
15
28
3kk
46
54

If kk is a positive integer and the median processing time for all requests in the batch is equal to 3 days, what is the minimum possible value of kk?

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Cevap: 4

Cevap

4
The correct answer is 4. There are 5+8=135 + 8 = 13 values less than 33. For 33 to be the median of the dataset, the median position must be greater than 1313. With k=4k = 4, the total number of items is N=23+4=27N = 23 + 4 = 27. For an odd dataset of size 2727, the median is the 27+12=14\frac{27 + 1}{2} = 14 th item. Since the first 1313 items are less than 33, the 1414 th item is 33, making 33 the median. Any value of k3k \le 3 results in a median less than 33.

Adım Adım Çözüm

1
Calculate the cumulative frequency of values below 3 days.
Number of requests with processing times of 1 or 2 days is 5+8=135 + 8 = 13.
To determine the median position, we first count how many data points lie strictly below 3 days.
2
Express the total number of requests NN in terms of kk.
N=5+8+k+6+4=23+kN = 5 + 8 + k + 6 + 4 = 23 + k.
The total number of requests determines whether NN is odd or even and where the median position lies.
3
Determine the condition for 3 days to be the median.
For 3 days to be the median, the median position must be greater than 13 so that it falls into the category of 3 days.
Since 13 values are strictly less than 3, the median rank must be at least 14.
4
Test minimum integer values for kk.
If k=3k = 3, N=26N = 26. The median is the average of the 1313 th value (22) and the 1414 th value (33), which is 2.52.5. If k=4k = 4, N=27N = 27. The median is the 1414 th value, which is 33.
Testing k=4k = 4 yields N=27N = 27 where the 1414 th value is 33, satisfying the median requirement.

Anahtar Kavram

Median of a Frequency Distribution Table
Tahmini Süre:1m 30s
Soru 3Soru

A dataset SS consists of 99 distinct integers. The median of SS is 2020, and the mean of SS is 2424. The mean of the 44 smallest integers in SS is 1010. If the largest integer in SS is removed to form a new dataset RR consisting of 88 integers, what is the maximum possible median of dataset RR?

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Cevap: 19.519.5

Cevap

The maximum possible median of dataset RR is 19.519.5.
In the sorted set of 99 distinct integers, the median is the 55 th element, which is 2020. Removing the largest value leaves 88 integers, whose median is the average of the 44 th and 55 th elements. Because all integers are distinct, the 44 th element can be at most 1919. The maximum median is therefore 19+202=19.5\frac{19 + 20}{2} = 19.5.

Adım Adım Çözüm

1
Identify the position of the median in the original dataset SS.
In a dataset of 99 sorted distinct integers x1<x2<x3<x4<x5<x6<x7<x8<x9x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9, the median is the 55 th element, so x5=20x_5 = 20.
For an odd number of ordered elements n=9n=9, the median is at index 9+12=5\frac{9+1}{2} = 5.
2
Determine the formula for the median of the modified dataset RR.
Removing the largest element x9x_9 leaves 88 ordered elements x1,x2,x3,x4,x5,x6,x7,x8x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8. The median of RR is x4+x52=x4+202\frac{x_4 + x_5}{2} = \frac{x_4 + 20}{2}.
For an even number of ordered elements n=8n=8, the median is the average of the two middle terms at indices 44 and 55.
3
Maximize the value of x4x_4.
Since all integers are distinct and x5=20x_5 = 20, x4x_4 must be an integer strictly less than 2020. Thus, the maximum integer value x4x_4 can take is 1919.
Maximizing x4x_4 maximizes the average x4+202\frac{x_4 + 20}{2}.
4
Verify that x4=19x_4 = 19 is achievable under the given mean constraint.
The sum of the smallest 44 integers is 4×10=404 \times 10 = 40. With x4=19x_4 = 19, we require x1+x2+x3=21x_1 + x_2 + x_3 = 21. Choosing distinct integers such as x1=1x_1 = 1, x2=2x_2 = 2, and x3=18x_3 = 18 satisfies 1+2+18=211 + 2 + 18 = 21 and x1<x2<x3<x4x_1 < x_2 < x_3 < x_4.
Confirming feasibility ensures the upper bound 1919 is valid.
5
Calculate the maximum median of dataset RR.
19+202=19.5\frac{19 + 20}{2} = 19.5.
Substitute the maximum value of x4=19x_4 = 19 into the median expression.

Anahtar Kavram

Properties of median for odd vs. even datasets and optimization under distinctness constraints
Soru 4Soru

An environmental monitoring station recorded the daily nitrogen dioxide (NO2\text{NO}_2) concentrations, in parts per billion (ppb), for 6 consecutive days: 19,28,30,35,42,19, 28, 30, 35, 42, and 4242. On the 7th day, a measurement of xx ppb was recorded, where x>45x > 45. If the arithmetic mean of the 7 daily concentrations is equal to their median, what is the value of xx?

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Cevap: 49

Cevap

The value of xx is 4949.
Sorting the first 6 recorded concentrations gives 19,28,30,35,42,4219, 28, 30, 35, 42, 42. Because x>45x > 45, xx is strictly greater than all existing values, so the full set in ascending order is 19,28,30,35,42,42,x19, 28, 30, 35, 42, 42, x. The median of a 7-element set is the 4th value, which is 3535. The sum of the 7 concentrations is 19+28+30+35+42+42+x=196+x19 + 28 + 30 + 35 + 42 + 42 + x = 196 + x, making the arithmetic mean 196+x7\frac{196 + x}{7}. Setting the mean equal to the median gives 196+x7=35    196+x=245    x=49\frac{196 + x}{7} = 35 \implies 196 + x = 245 \implies x = 49.

Adım Adım Çözüm

1
Order the first 6 data points in ascending order
The sorted list is 19,28,30,35,42,4219, 28, 30, 35, 42, 42.
Establishing ordered positions is necessary to determine the median.
2
Determine the median of the 7-element dataset
Since x>45x > 45, the complete ordered dataset is 19,28,30,35,42,42,x19, 28, 30, 35, 42, 42, x, making the 4th element, 3535, the median.
The median of an odd number of sorted values is the middle element.
3
Express the arithmetic mean in terms of xx and set it equal to the median
196+x7=35\frac{196 + x}{7} = 35
The problem states that the arithmetic mean equals the median.
4
Solve the linear equation for xx
196 + x = 245 \implies x = 49
Multiplying both sides by 7 and subtracting 196 isolates xx.

Anahtar Kavram

Measures of Central Tendency (Mean and Median)
Soru 5Soru

A specialty tea shop sells five distinct sizes of loose-leaf tea canisters. The table below shows the weight per canister and the number of canisters sold during a promotional week:

Canister SizeWeight per Canister (ounces)Number of Canisters Sold
Sampler516
Small1218
Medium15nn
Large206
Executive254

If the mean weight of all canisters sold during the week is 1313 ounces, what is the median weight, in ounces, of the canisters sold?

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Cevap: 1515 ounces

Cevap

15 ounces
The correct answer is 15 ounces. Calculating the total weight as a function of nn gives 516+15n516 + 15n, and the total number of canisters as 44+n44 + n. Setting the mean 516+15n44+n=13\frac{516 + 15n}{44 + n} = 13 yields n=28n = 28. The total number of canisters sold is 44+28=7244 + 28 = 72. For 72 items sorted in ascending order, the median is the average of the 36th and 37th values. Cumulative frequencies show that the 1st through 16th items are 5 ounces, the 17th through 34th items are 12 ounces, and the 35th through 62nd items are 15 ounces. Thus, both the 36th and 37th items weigh 15 ounces, making the median 15 ounces.

Adım Adım Çözüm

1
Set up an equation for the total number of canisters and total weight.
Total canisters = 16+18+n+6+4=44+n16 + 18 + n + 6 + 4 = 44 + n. Total weight = 5(16)+12(18)+15(n)+20(6)+25(4)=516+15n5(16) + 12(18) + 15(n) + 20(6) + 25(4) = 516 + 15n ounces.
The mean of a weighted dataset is the total sum of all values divided by the total number of items.
2
Solve for the missing frequency nn using the given mean of 13 ounces.
516+15n44+n=13    516+15n=13(44+n)    516+15n=572+13n    2n=56    n=28\frac{516 + 15n}{44 + n} = 13 \implies 516 + 15n = 13(44 + n) \implies 516 + 15n = 572 + 13n \implies 2n = 56 \implies n = 28.
Equating the weighted average expression to 13 allows us to determine the exact number of Medium canisters sold.
3
Determine the total number of items and locate the position of the median.
Total canisters = 44+28=7244 + 28 = 72. The median is the average of the 36th36^{\text{th}} and 37th37^{\text{th}} values when ordered.
For an even number of data points N=72N = 72, the median is the arithmetic mean of the (N/2)th(N/2)^{\text{th}} and (N/2+1)th(N/2 + 1)^{\text{th}} items.
4
Compute the cumulative frequency distribution to find the values at the 36th36^{\text{th}} and 37th37^{\text{th}} positions.
5 oz: positions 1 to 16; 12 oz: positions 17 to 34; 15 oz: positions 35 to 62. Both 36th36^{\text{th}} and 37th37^{\text{th}} positions have a weight of 15 ounces, so the median is 15 ounces.
Cumulative counting reveals that items in positions 35 through 62 all correspond to 15-ounce canisters.

Anahtar Kavram

Weighted mean and finding the median of frequency distribution data
Soru 6Soru

The frequency distribution table below summarizes the scores of 20 students on a statistics test.

ScoreNumber of Students
53
65
7xx
8yy
94

If the arithmetic mean score of the 20 students is 7.157.15, what is the median score of the 20 students?

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Cevap: 7.5

Cevap

The median score of the 20 students is 7.5.
To find the median score of 20 students, we first determine the missing student counts xx and yy. From the total count of 20 students, x+y=20(3+5+4)=8x + y = 20 - (3 + 5 + 4) = 8. Using the given mean of 7.15, the sum of all scores is 20×7.15=14320 \times 7.15 = 143. The sum expressed in terms of xx and yy is 3(5)+5(6)+7x+8y+4(9)=81+7x+8y3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y. Setting 81+7x+8y=14381 + 7x + 8y = 143 gives 7x+8y=627x + 8y = 62. Solving x+y=8x + y = 8 and 7x+8y=627x + 8y = 62 yields x=2x = 2 and y=6y = 6. Listing scores in non-decreasing order, the 10th student scored 7 and the 11th student scored 8. Because the total number of students is even, the median is the average of these two middle values: (7+8)/2=7.5(7 + 8) / 2 = 7.5.

Adım Adım Çözüm

1
Set up an equation for the total number of students to find the sum of xx and yy.
3+5+x+y+4=20    x+y+12=20    x+y=83 + 5 + x + y + 4 = 20 \implies x + y + 12 = 20 \implies x + y = 8.
The total frequency of all score categories must equal the total number of students (20).
2
Set up an equation for the sum of all scores using the given arithmetic mean.
Total sum =20×7.15=143= 20 \times 7.15 = 143. Also, total sum =3(5)+5(6)+7x+8y+4(9)=81+7x+8y= 3(5) + 5(6) + 7x + 8y + 4(9) = 81 + 7x + 8y. Therefore, 81+7x+8y=143    7x+8y=6281 + 7x + 8y = 143 \implies 7x + 8y = 62.
The sum of all data values equals the mean multiplied by the total number of observations.
3
Solve the system of linear equations for xx and yy.
Multiply x+y=8x + y = 8 by 7 to get 7x+7y=567x + 7y = 56. Subtract this from 7x+8y=627x + 8y = 62 to get y=6y = 6. Then x=86=2x = 8 - 6 = 2.
Determining the exact frequencies x=2x=2 and y=6y=6 is required to locate the middle values.
4
Determine the cumulative frequencies and find the 10th and 11th values.
Score 5: positions 1 to 3 (3 students); Score 6: positions 4 to 8 (5 students); Score 7: positions 9 to 10 (2 students); Score 8: positions 11 to 16 (6 students); Score 9: positions 17 to 20 (4 students). The 10th value is 7 and the 11th value is 8.
For an even number of data points (N=20N=20), the median is the arithmetic mean of the (N/2)(N/2)-th and (N/2+1)(N/2 + 1)-th values.
5
Calculate the median score.
Median=7+82=7.5\text{Median} = \frac{7 + 8}{2} = 7.5.
Average the two middle values.

Anahtar Kavram

Calculating the median from a frequency distribution table with missing counts by combining system of linear equations with cumulative frequency analysis.
Soru 7Soru

A store manager recorded the number of orders processed online each day during a one-week period: 1414, 77, 2525, 1111, 1818, 22, and 1616. What is the median number of daily online orders processed during this week?

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Cevap: 1414

Cevap

The median number of daily online orders is 1414.
To find the median, the seven numbers must first be placed in ascending order: 2,7,11,14,16,18,252, 7, 11, 14, 16, 18, 25. Since there is an odd number of data points (77), the median is the single middle value located at the 4th position, which is 1414.

Adım Adım Çözüm

1
Arrange the given dataset of seven numbers in ascending order.
The sorted dataset is 2,7,11,14,16,18,252, 7, 11, 14, 16, 18, 25.
To determine the median of any dataset, the numerical values must first be ordered from smallest to largest.
2
Find the position of the middle element for a dataset with n=7n = 7 values.
The median position is 7+12=4\frac{7 + 1}{2} = 4.
When a dataset contains an odd number of elements nn, the median is located at the n+12\frac{n+1}{2}-th position.
3
Identify the 4th element in the sorted list.
The 4th element is 1414.
In the ordered sequence 2,7,11,14,16,18,252, 7, 11, 14, 16, 18, 25, the 4th entry is 1414.

Anahtar Kavram

The median is the middle value in a set of numerical data that has been arranged in ascending or descending order.
Soru 8Soru

A researcher recorded seven daily temperature readings (in degrees Celsius): 14,5,22,9,25,2,14, 5, 22, 9, 25, 2, and 1717. What is the median of these seven temperature readings?

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Cevap: 1414

Cevap

The median of the temperature readings is 1414.
To find the median of a set of numbers, first arrange them in ascending order: 2,5,9,14,17,22,252, 5, 9, 14, 17, 22, 25. Since there are 77 numbers, the median is the middle value in the 4th4\text{th} position, which is 1414.

Adım Adım Çözüm

1
Arrange the given dataset in ascending numerical order.
The ordered list of 77 temperatures is: 2,5,9,14,17,22,252, 5, 9, 14, 17, 22, 25.
To calculate the median of a dataset, the values must first be ordered from smallest to largest.
2
Identify the middle position for an odd number of data values (n=7n = 7).
The position of the median is 7+12=4th\frac{7 + 1}{2} = 4\text{th} position.
For a dataset with nn elements where nn is odd, the median is located at position n+12\frac{n+1}{2}.
3
Extract the value located at the 4th4\text{th} position in the ordered list.
The 4th4\text{th} term is 1414.
The fourth number in 2,5,9,14,17,22,252, 5, 9, 14, 17, 22, 25 is 1414.

Anahtar Kavram

Median of a Dataset
Tahmini Süre:45s
Soru 9Soru

A customer service representative resolved the following number of support tickets over five consecutive days: 1818, 2424, 1515, 3131, and 2222. What is the arithmetic mean of the number of tickets resolved per day by the representative?

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Cevap: 22

Cevap

The arithmetic mean of the number of tickets resolved per day is 2222.
The mean of a data set is calculated by taking the sum of all values and dividing by the total number of items. The sum of the numbers is 18+24+15+31+22=11018 + 24 + 15 + 31 + 22 = 110. Dividing 110110 by 55 yields 2222.

Adım Adım Çözüm

1
Sum all data values in the set
18+24+15+31+22=11018 + 24 + 15 + 31 + 22 = 110
To find the mean, the first step is to calculate the total sum of all observations.
2
Divide the total sum by the total number of values
1105=22\frac{110}{5} = 22
The arithmetic mean is defined as the sum of the values divided by the count of the values.

Anahtar Kavram

Arithmetic Mean
Soru 10Soru

A high school basketball team scored the following points in five consecutive games: 44, 66, 66, 77, and 1212. Which of the following statements regarding the measures of central tendency for these scores must be true? Select all such statements.

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Cevap: The mean score is equal to 77.; The median score is equal to the mode score.; The mode score is equal to 66.

Cevap

The statements asserting that the mean score is equal to 7, the median score is equal to the mode score, and the mode score is equal to 6 are all true.
For the dataset 4,6,6,7,124, 6, 6, 7, 12, the mean is 355=7\frac{35}{5} = 7, the median (middle score) is 66, and the mode (most frequent score) is 66. Consequently, the mean equals 77, the median and mode are equal (6=66 = 6), and the mode is 66.

Adım Adım Çözüm

1
Calculate the mean of the dataset.
Mean = 4+6+6+7+125=355=7\frac{4 + 6 + 6 + 7 + 12}{5} = \frac{35}{5} = 7.
The mean is calculated by dividing the sum of all data points by the total count.
2
Determine the median of the dataset.
Median = 66.
The data is already arranged in ascending order: 4,6,6,7,124, 6, 6, 7, 12. The 3rd value (middle position) is 66.
3
Determine the mode of the dataset.
Mode = 66.
The value 66 occurs twice, whereas all other values occur only once.
4
Evaluate each given statement against the calculated metrics.
Mean = 77, Median = 66, Mode = 66. Thus: Mean (77) is true; Median = Mode (6=66 = 6) is true; Mode = 66 is true.
Direct comparison verifies which statements hold true.

Anahtar Kavram

Calculating and comparing mean, median, and mode for a discrete set of numerical data.
Tahmini Süre:1m 0s
Soru 11Soru

A company has two sales divisions, Division X and Division Y. Division X has 12 representatives with an average monthly sales volume of $14,000\$14,000. Division Y has 18 representatives with an average monthly sales volume of $24,000\$24,000. If 2 representatives leave Division Y who together accounted for $82,000\$82,000 in monthly sales, what is the new combined average monthly sales volume, in dollars, for all remaining representatives in both divisions?

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Cevap: 18500

Cevap

The new combined average monthly sales volume for all remaining representatives is $18,500.
To find the combined mean, calculate the sum of all remaining values across both groups (518,000518,000) and divide by the total number of remaining items (2828), yielding 18,50018,500.

Adım Adım Çözüm

1
Find total monthly sales for Division X
12×14,000=168,00012 \times 14,000 = 168,000
The total value of a group is the product of its count and its arithmetic mean.
2
Find total monthly sales for Division Y before representatives left
18×24,000=432,00018 \times 24,000 = 432,000
Multiply the number of representatives in Division Y by their average sales volume.
3
Determine the remaining total sales in Division Y
432,00082,000=350,000432,000 - 82,000 = 350,000
Subtract the sales volume of the 2 departing representatives from Division Y's original total.
4
Calculate the total combined sales for both divisions
168,000+350,000=518,000168,000 + 350,000 = 518,000
Add the total sales of Division X and the remaining sales of Division Y.
5
Calculate total number of remaining representatives across both divisions
12 + (18 - 2) = 28
Division X retains 12 representatives, while Division Y retains 16 representatives.
6
Compute the combined mean sales volume
518,00028=18,500\frac{518,000}{28} = 18,500
Divide the total combined sales volume by the total number of remaining representatives.

Anahtar Kavram

Weighted Mean and Combining Group Data
Soru 12Soru

A dataset consists of 99 distinct positive integers a1,a2,a3,a4,a5,a6,a7,a8,a9a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9 listed in strictly increasing order. The mean of the 99 integers is 2828, and the median is 2424. A 10th10\text{th} positive integer xx, where x>a9x > a_9, is added to the dataset, causing the new mean of the 1010 integers to become 3131. If LL represents the minimum possible value of a9a_9, what is the value of xLx - L?

Cevabı ve açıklamayı göster

Cevap: 2121

Cevap

The value of xLx - L is 2121.
The sum of the original 99 elements is 9×28=2529 \times 28 = 252, and the sum of the 1010 elements is 10×31=31010 \times 31 = 310, giving x=58x = 58. To minimize a9a_9 (LL), the sum of all other elements must be maximized. The median a5=24a_5 = 24. The maximum possible values for the first four distinct elements below 2424 are 20,21,22,2320, 21, 22, 23 (summing to 8686). To minimize a9a_9, the upper four elements must be consecutive integers (a93,a92,a91,a9)(a_9 - 3, a_9 - 2, a_9 - 1, a_9). Setting the total sum equation 86+24+(4a96)=25286 + 24 + (4a_9 - 6) = 252 yields 4a9=1484a_9 = 148, so L=37L = 37. Consequently, xL=5837=21x - L = 58 - 37 = 21.

Adım Adım Çözüm

1
Calculate the sum of the original 9 integers and determine the value of the 10th integer xx.
The sum of the original 99 integers is 9×28=2529 \times 28 = 252. The sum of the 1010 integers after adding xx is 10×31=31010 \times 31 = 310. Therefore, x=310252=58x = 310 - 252 = 58.
The sum of a set of numbers equals the number of elements multiplied by the mean.
2
Identify the median of the ordered 9-element set.
In an ordered set of 99 elements, the median is the 5th5\text{th} element, so a5=24a_5 = 24.
For an odd number of ordered elements, the median is the exact middle element.
3
Maximize the sum of the first 4 elements a1,a2,a3,a4a_1, a_2, a_3, a_4 to minimize a9a_9.
Since all integers are distinct and strictly increasing, a4<24a_4 < 24. To maximize a1+a2+a3+a4a_1 + a_2 + a_3 + a_4, choose a4=23,a3=22,a2=21,a1=20a_4 = 23, a_3 = 22, a_2 = 21, a_1 = 20. Their sum is 20+21+22+23=8620 + 21 + 22 + 23 = 86.
Maximizing lower elements leaves the smallest possible remainder of the total sum for the upper elements.
4
Express a6,a7,a8a_6, a_7, a_8 in terms of a9a_9 to minimize a9a_9.
To make a9a_9 as small as possible, a6,a7,a8a_6, a_7, a_8 should be as large as possible relative to a9a_9, meaning they are consecutive integers: a8=a91a_8 = a_9 - 1, a7=a92a_7 = a_9 - 2, a6=a93a_6 = a_9 - 3.
Making elements above the median consecutive integers directly below a9a_9 minimizes a9a_9 for a fixed sum.
5
Set up the sum equation for the 9 elements to solve for L=min(a9)L = \text{min}(a_9).
(a1+a2+a3+a4)+a5+(a6+a7+a8+a9)=252    86+24+(a93+a92+a91+a9)=252    104+4a9=252    4a9=148    a9=37(a_1 + a_2 + a_3 + a_4) + a_5 + (a_6 + a_7 + a_8 + a_9) = 252 \implies 86 + 24 + (a_9 - 3 + a_9 - 2 + a_9 - 1 + a_9) = 252 \implies 104 + 4a_9 = 252 \implies 4a_9 = 148 \implies a_9 = 37. Thus, L=37L = 37.
Solving the algebraic equation derived from the total sum yields the minimum integer value for a9a_9 while satisfying a6=34>24a_6 = 34 > 24.
6
Calculate xLx - L.
xL=5837=21x - L = 58 - 37 = 21.
Subtracting the minimum bound LL from xx fulfills the target question requirement.

Anahtar Kavram

Optimization of Dataset Values using Central Tendency Constraints
Tahmini Süre:2m 30s
Soru 13Soru

A department consisting of 55 employees has a mean monthly sales total of $12000\$12{}000. If a new employee with a monthly sales total of $18000\$18{}000 joins the department, what is the new mean monthly sales total, in dollars, for the 66 employees?

Cevabı ve açıklamayı göster

Cevap: 13000

Cevap

13000
To find the new mean, multiply the initial mean by the initial number of employees to get total sales: 5×12000=600005 \times 12{}000 = 60{}000. Add the new employee's sales: 60000+18000=7800060{}000 + 18{}000 = 78{}000. Finally, divide by the new total number of employees (66) to get 780006=13000\frac{78{}000}{6} = 13{}000.

Adım Adım Çözüm

1
Find the total sales of the original 5 employees.
5×12000=600005 \times 12{}000 = 60{}000
The sum of values is equal to the mean multiplied by the number of observations.
2
Calculate the total sales for all 6 employees.
60000+18000=7800060{}000 + 18{}000 = 78{}000
Add the new employee's sales to the initial total.
3
Calculate the new mean sales per employee.
780006=13000\frac{78{}000}{6} = 13{}000
Divide the combined total sales by the new total number of employees (6).

Anahtar Kavram

Mean of Combined Data Sets
Tahmini Süre:1m 0s
Soru 14Soru

A dataset consists of 2525 distinct integers arranged in increasing order. The mean of all 2525 integers is 5252. The mean of the smallest 1212 integers is 3030, and the mean of the largest 1212 integers is 7070.

If 55 additional numbers, each equal to the median of the original dataset, are added to the dataset, what is the mean of the new set of 3030 numbers?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

The mean of the new set of 30 numbers is 60.
The sum of all 25 numbers is 25×52=130025 \times 52 = 1300. Since the numbers are ordered, the median is the 13th value. The 12 smallest numbers sum to 12×30=36012 \times 30 = 360 and the 12 largest sum to 12×70=84012 \times 70 = 840. The sum of these 24 numbers is 360+840=1200360 + 840 = 1200, so the 13th number (the median) must be 13001200=1001300 - 1200 = 100. Adding 5 copies of 100 increases the sum to 1300+500=18001300 + 500 = 1800 across 3030 numbers. The new mean is 180030=60\frac{1800}{30} = 60.

Adım Adım Çözüm

1
Calculate the total sum of the original dataset of 25 numbers.
Sum = 25×52=130025 \times 52 = 1300.
The mean multiplied by the number of elements gives the total sum.
2
Calculate the combined sum of the 12 smallest and 12 largest integers.
Sum of 12 smallest = 12×30=36012 \times 30 = 360; Sum of 12 largest = 12×70=84012 \times 70 = 840; Total = 360+840=1200360 + 840 = 1200.
The 25 numbers consist of the 12 smallest, the 1 median (13th element), and the 12 largest.
3
Determine the value of the median.
Median = 13001200=1001300 - 1200 = 100.
Subtracting the sum of the 24 non-median values from the total sum yields the 13th element, which is the median.
4
Find the sum and count of the modified dataset.
New Sum = 1300+5(100)=18001300 + 5(100) = 1800; New Count = 25+5=3025 + 5 = 30.
Adding 5 numbers each equal to 100 increases the sum by 500 and the count by 5.
5
Calculate the mean of the new dataset.
New Mean = 180030=60\frac{1800}{30} = 60.
Divide the new total sum by the new total count of numbers.

Anahtar Kavram

Relationship between Mean, Median, and Total Sum in Partitioned Datasets
Tahmini Süre:2m 30s
Soru 15Soru

A software engineer logged the response time, in milliseconds, for seven independent server requests: 88, 1616, 33, 2121, 1414, 55, and 1010. What is the median response time, in milliseconds, for these seven requests?

Cevabı ve açıklamayı göster

Cevap: 1010

Cevap

The median response time is 1010 milliseconds.
To find the median, first order the seven data points in ascending order: 3,5,8,10,14,16,213, 5, 8, 10, 14, 16, 21. Since there are 77 values, the median is the middle value at position 7+12=4\frac{7 + 1}{2} = 4. The fourth value in this ordered list is 1010.

Adım Adım Çözüm

1
Arrange the given dataset in ascending order.
The sorted dataset is 3,5,8,10,14,16,213, 5, 8, 10, 14, 16, 21.
Finding the median of a numerical dataset requires ordering the values from smallest to largest first.
2
Determine the position of the median element.
For n=7n = 7 items, the position is 7+12=4\frac{7 + 1}{2} = 4 th item.
When the number of observations nn is odd, the median is the exact middle value located at position n+12\frac{n+1}{2}.
3
Identify the value at the 4th position.
The 4th value in the sorted list is 1010.
The 4th element in 3,5,8,10,14,16,213, 5, 8, 10, 14, 16, 21 is 1010.

Anahtar Kavram

Median of a finite numerical dataset with an odd count
Tahmini Süre:45s
Soru 16Soru

Group A consists of 1515 test scores with an arithmetic mean of 8080 and a median of 8282. Group B consists of 2525 test scores with an arithmetic mean of 9090 and a median of 8686. The two groups are combined into a single dataset of 4040 test scores. Which of the following statements MUST be true about the combined dataset? Indicate all such statements.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: The mean of the combined dataset is equal to 86.2586.25.; The median of the combined dataset is at least 8282.; The median of the combined dataset is at most 8686.

Cevap

The statements asserting that the combined mean is 86.2586.25, that the combined median is at least 8282, and that the combined median is at most 8686 must all be true.
The combined mean is calculated directly from the total sum divided by total elements, giving exactly 86.2586.25. For the median of 4040 combined values (the average of the 20th20\text{th} and 21st21\text{st} values in sorted order), counting how many values are guaranteed to be above or below specific thresholds shows that at least 2121 values are 82\ge 82 and at least 2121 values are 86\le 86. Therefore, the median must lie within [82,86][82, 86].

Adım Adım Çözüm

1
Calculate the weighted arithmetic mean for the combined dataset.
Combined Mean = 15×80+25×9015+25=1200+225040=345040=86.25\frac{15 \times 80 + 25 \times 90}{15 + 25} = \frac{1200 + 2250}{40} = \frac{3450}{40} = 86.25.
The mean of combined sets is the total sum of all values divided by the total number of items.
2
Analyze the lower bound for the median of the combined set.
At least 2121 values in the combined set of 4040 items are greater than or equal to 8282.
Since the median of Group A (1515 items) is 8282, at least 88 items are 82\ge 82. Since the median of Group B (2525 items) is 8686, at least 1313 items are 8682\ge 86 \ge 82. In total, 8+13=218 + 13 = 21 items are 82\ge 82. In a sorted list of 4040 items, having at least 2121 items 82\ge 82 implies that the 20th20\text{th} and 21st21\text{st} items must be 82\ge 82, so the median 82\ge 82.
3
Analyze the upper bound for the median of the combined set.
At least 2121 values in the combined set of 4040 items are less than or equal to 8686.
Group A has at least 88 items 8286\le 82 \le 86. Group B has at least 1313 items 86\le 86. Thus, at least 8+13=218 + 13 = 21 items are 86\le 86. In a sorted list of 4040 items, having at least 2121 items 86\le 86 ensures the 20th20\text{th} and 21st21\text{st} items are 86\le 86, so the median 86\le 86.
4
Evaluate the remaining statements regarding the exact median and mode.
Neither the exact median value of 8484 nor the existence/location of a mode is guaranteed.
Medians do not combine linearly like means, and mode depends entirely on specific frequency counts which are not given.

Anahtar Kavram

Weighted averages and order statistic bounds for combined datasets
Soru 17Soru

A dataset DD consists of 1111 distinct positive integers. The median of DD is 4040, and the arithmetic mean of DD is 4545. A new dataset DD' is created by increasing each of the 55 largest integers in DD by 1010 and decreasing each of the 55 smallest integers in DD by a positive integer xx. If the median of DD' is strictly less than the arithmetic mean of DD', what is the maximum possible integer value of xx?

Cevabı ve açıklamayı göster

Cevap: 2020

Cevap

The maximum possible integer value of xx is 2020.
The sum of the original 1111 values is 11×45=49511 \times 45 = 495. When the 55 largest integers are each increased by 1010, the sum increases by +50+50. When the 55 smallest integers are each decreased by xx, the sum decreases by 5x-5x. The new sum is 5455x545 - 5x, making the new mean 5455x11\frac{545 - 5x}{11}. Because DD has 1111 elements, the median is the 6th6^{\text{th}} element. Changing the smallest 55 and largest 55 elements does not alter the value of the 6th6^{\text{th}} element, so the median remains 4040. Requiring the median to be strictly less than the new mean gives 40<5455x11    440<5455x    5x<105    x<2140 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21. The greatest integer less than 2121 is 2020.

Adım Adım Çözüm

1
Calculate the total sum of the original dataset DD.
Since DD has 1111 elements with a mean of 4545, the sum of elements is 11×45=49511 \times 45 = 495.
The mean formula is Mean=Sumn\text{Mean} = \frac{\text{Sum}}{n}, so Sum=n×Mean\text{Sum} = n \times \text{Mean}.
2
Determine the median and sum of the modified dataset DD'.
The median remains the 6th6^{\text{th}} element, which is 4040. The sum of DD' is 495+5(10)5(x)=5455x495 + 5(10) - 5(x) = 545 - 5x.
Modifying only the 55 smallest and 55 largest elements leaves the 6th6^{\text{th}} central element unchanged.
3
Set up and solve the inequality comparing the median to the mean of DD'.
Solve 40<5455x11    440<5455x    5x<105    x<2140 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21.
The problem specifies that the median must be strictly less than the mean.
4
Identify the maximum integer value satisfying the inequality.
The largest integer strictly less than 2121 is 2020.
xx must be an integer.

Anahtar Kavram

Effect of data modifications on mean and median
Tahmini Süre:2m 30s
Soru 18Soru

A beverage manufacturing company tracks the monthly production counts (in thousands of units) for its facilities located in two separate regions. Region A consists of 66 facilities with monthly production counts of 1212, 1515, 1515, 1818, 2020, and 2828. Region B consists of 44 facilities with monthly production counts of 1010, 1414, 2222, and 3434.

Which of the following statements must be true regarding the individual regional datasets and the combined dataset of all 1010 facilities? Select all such statements.

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Cevabı ve açıklamayı göster

Cevap: The arithmetic mean of the combined dataset of all 1010 facilities is greater than the arithmetic mean of Region A.; The median of the combined dataset of all 1010 facilities is equal to the median of Region A.

Cevap

The correct statements are the one asserting that the combined arithmetic mean is greater than Region A's mean, and the one asserting that the combined median is equal to Region A's median.
The mean of the combined dataset (18.818.8) is greater than the mean of Region A (18.018.0), making the statement comparing the combined mean to Region A's mean correct. Additionally, both Region A's median and the combined dataset's median evaluate to 16.516.5, making the statement asserting equality between these two medians correct.

Adım Adım Çözüm

1
Calculate the sum, mean, and median for Region A.
Region A sum =12+15+15+18+20+28=108= 12 + 15 + 15 + 18 + 20 + 28 = 108. Mean =1086=18= \frac{108}{6} = 18. Sorted values are 12,15,15,18,20,2812, 15, 15, 18, 20, 28, so Median =15+182=16.5= \frac{15 + 18}{2} = 16.5. Mode =15= 15.
Establishing base central metrics for the first subgroup.
2
Calculate the sum, mean, and median for Region B.
Region B sum =10+14+22+34=80= 10 + 14 + 22 + 34 = 80. Mean =804=20= \frac{80}{4} = 20. Sorted values are 10,14,22,3410, 14, 22, 34, so Median =14+222=18= \frac{14 + 22}{2} = 18.
Establishing base central metrics for the second subgroup.
3
Combine and order all 10 values to calculate combined metrics.
Combined ordered set: 10,12,14,15,15,18,20,22,28,3410, 12, 14, 15, 15, 18, 20, 22, 28, 34. Total sum =108+80=188= 108 + 80 = 188. Combined Mean =18810=18.8= \frac{188}{10} = 18.8. Combined Median =15+182=16.5= \frac{15 + 18}{2} = 16.5. Combined Mode =15= 15.
Necessary to evaluate combined properties accurately against subgroup properties.
4
Evaluate each candidate statement against calculated values.
Combined mean (18.818.8) > Region A mean (1818) is TRUE. Combined median (16.516.5) = Region A median (16.516.5) is TRUE. Average of medians (17.2517.25) = Combined median (16.516.5) is FALSE. Combined mode (1515) > Region A mode (1515) is FALSE. Simple average of means (1919) = Combined mean (18.818.8) is FALSE.
Determining which statements satisfy the required conditions.

Anahtar Kavram

Combining Datasets and Weighted Measures of Central Tendency
Soru 19Soru

A dataset SS consists of 77 distinct positive integers arranged in ascending order: x1,x2,x3,x4,x5,x6,x7x_1, x_2, x_3, x_4, x_5, x_6, x_7. The arithmetic mean of the entire dataset is 2828, and the median is 2525. The arithmetic mean of the 33 smallest integers in SS is 1212. If MM is the maximum possible value of x7x_7 and mm is the minimum possible value of x7x_7, what is the value of MmM - m?

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

The value of MmM - m is 3636.
The sum of all 77 distinct positive integers is 7×28=1967 \times 28 = 196. Since the dataset is ordered and has 77 elements, the median is x4=25x_4 = 25. The sum of the smallest 33 integers is 3×12=363 \times 12 = 36. Therefore, the sum of the remaining three integers x5+x6+x7=1963625=135x_5 + x_6 + x_7 = 196 - 36 - 25 = 135.

To maximize x7x_7, x5x_5 and x6x_6 must be as small as possible. Since all elements are distinct integers greater than x4=25x_4 = 25, the smallest possible values are x5=26x_5 = 26 and x6=27x_6 = 27. Thus, M=1352627=82M = 135 - 26 - 27 = 82.

To minimize x7x_7, x5x_5, x6x_6, and x7x_7 must be as close together as possible while preserving 25<x5<x6<x725 < x_5 < x_6 < x_7. Dividing 135135 by 33 gives 4545. The consecutive integers centered around 4545 are 44,45,4644, 45, 46, which sum to 135135 and satisfy all inequalities. Thus, m=46m = 46.

The difference Mm=8246=36M - m = 82 - 46 = 36, which corresponds to the value 3636.

Adım Adım Çözüm

1
Calculate the total sum of all 7 integers in dataset S.
Total sum = 7×28=1967 \times 28 = 196.
The arithmetic mean of nn numbers is the total sum divided by nn.
2
Identify the median value and the sum of the smallest 3 integers.
Median x4=25x_4 = 25, and x1+x2+x3=3×12=36x_1 + x_2 + x_3 = 3 \times 12 = 36.
For an odd number of ordered elements (77), the middle term x4x_4 is the median. The mean of the first 3 terms gives their sum.
3
Determine the sum of the top 3 integers (x5+x6+x7)(x_5 + x_6 + x_7).
x5+x6+x7=1963625=135x_5 + x_6 + x_7 = 196 - 36 - 25 = 135.
Subtracting x1+x2+x3x_1 + x_2 + x_3 and x4x_4 from the total sum leaves the sum of the remaining three elements.
4
Calculate the maximum possible value MM of x7x_7.
M=82M = 82.
To maximize x7x_7, minimize x5x_5 and x6x_6. Since elements are distinct integers and x4=25x_4 = 25, the minimum values are x5=26x_5 = 26 and x6=27x_6 = 27. Thus x7=1352627=82x_7 = 135 - 26 - 27 = 82.
5
Calculate the minimum possible value mm of x7x_7.
m=46m = 46.
To minimize x7x_7, maximize x5x_5 and x6x_6 such that 25<x5<x6<x725 < x_5 < x_6 < x_7 and x5+x6+x7=135x_5 + x_6 + x_7 = 135. Setting x5=44,x6=45,x7=46x_5 = 44, x_6 = 45, x_7 = 46 gives 44+45+46=13544 + 45 + 46 = 135, maintaining strict inequalities.
6
Compute MmM - m.
Mm=8246=36M - m = 82 - 46 = 36.
Subtract the minimum possible value of x7x_7 from its maximum possible value.

Anahtar Kavram

Measures of Central Tendency with Extreme Value Optimization
Tahmini Süre:2m 30s
Soru 20Soru

A company has 8 departments. The dataset of the number of employees in these 8 departments has a median of 42, a range of 25, and a unique mode of 38, which appears exactly 3 times. If no department has more than 55 employees, what is the maximum possible arithmetic mean of the number of employees across all 8 departments?

Cevabı ve açıklamayı göster

Cevap: 44.25

Cevap

The maximum possible arithmetic mean of the number of employees across all 8 departments is 44.25.
To maximize the mean, the sum of the 8 department sizes must be maximized under all given constraints. Ordering the dataset as x1x2x3x4x5x6x7x8x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8, the median requirement gives x4+x5=84x_4 + x_5 = 84. Since no element exceeds 55 and the range is 25, x1x_1 cannot be 38 because 38+25=63>5538 + 25 = 63 > 55. Hence, the three 38s must be x2=x3=x4=38x_2 = x_3 = x_4 = 38, which forces x5=46x_5 = 46. To maximize the sum, x8x_8 is set to its maximum limit of 55, forcing x1=5525=30x_1 = 55 - 25 = 30. Next, x7x_7 is set to 55, and x6x_6 is set to 54 so that 55 appears only twice and 38 remains the unique mode. The maximum sum is 30+38+38+38+46+54+55+55=35430 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354, giving a maximum mean of 354/8=44.25354 / 8 = 44.25.

Adım Adım Çözüm

1
Order the dataset variables and define constraints.
Let the department sizes be ordered as x1x2x3x4x5x6x7x8x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6 \le x_7 \le x_8.
Arranging values in ascending order allows direct analysis of median, range, and mode bounds.
2
Use the median to form an equation for the middle two elements.
x4+x52=42    x4+x5=84\frac{x_4 + x_5}{2} = 42 \implies x_4 + x_5 = 84.
For n=8n=8 elements, the median is the arithmetic mean of the 4th and 5th terms.
3
Determine the exact position of the three occurrences of 38.
x2=x3=x4=38x_2 = x_3 = x_4 = 38, forcing x5=8438=46x_5 = 84 - 38 = 46.
If x1=38x_1 = 38, then x8=38+25=63x_8 = 38 + 25 = 63, exceeding the upper bound of 55. Thus 38 cannot start at x1x_1, so it must occupy x2,x3,x4x_2, x_3, x_4.
4
Maximize the remaining elements x1,x6,x7,x8x_1, x_6, x_7, x_8.
x8=55x_8 = 55, x1=30x_1 = 30, x7=55x_7 = 55, and x6=54x_6 = 54.
To maximize the sum, set x8=55x_8 = 55, which fixes x1=5525=30x_1 = 55 - 25 = 30. Set x7=55x_7 = 55. x6x_6 can be at most 54 because setting x6=55x_6 = 55 would give 55 a frequency of 3, violating the unique mode requirement.
5
Calculate the maximum sum and arithmetic mean.
Sum =30+38+38+38+46+54+55+55=354= 30 + 38 + 38 + 38 + 46 + 54 + 55 + 55 = 354; Mean =354/8=44.25= 354 / 8 = 44.25.
Dividing the maximum total sum of 354 by 8 gives the maximum possible arithmetic mean.

Anahtar Kavram

Optimization of Means Subject to Central Tendency and Range Constraints
Tahmini Süre:2m 30s
Sayfa 1 / 2Sonraki