Percentiles and Quartiles

10 soru

Soru 1Soru

In a corporate evaluation of 200200 employees, performance scores range from 00 to 100100. An employee with a score of 8484 scored strictly higher than 150150 employees and strictly lower than 4242 employees, while the remaining employees received a score of exactly 8484. If the percentile rank of a score is defined as the percentage of all scores strictly below it plus half the percentage of all scores equal to it, what is the percentile rank of a score of 8484?

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Cevap: 77th percentile

Cevap

77th percentile
The correct answer represents the relative position of a score of 84. Since 150 employees scored lower and 8 employees scored equal to 84, applying the percentile formula yields ((150 + 4) / 200) * 100 = 77th percentile.

Adım Adım Çözüm

1
Determine the number of employees scoring exactly 84.
Number of employees scoring 84 = 20015042=8200 - 150 - 42 = 8.
The total group size equals the sum of students scoring below 84, above 84, and equal to 84.
2
Calculate the effective count of scores at or below 84 under the given percentile definition.
Effective count = 150+0.5×8=154150 + 0.5 \times 8 = 154.
The definition specifies adding the count of strictly lower scores to half the count of tied scores.
3
Convert the effective count to a percentile rank.
Percentile rank = 154200×100=77th percentile\frac{154}{200} \times 100 = 77\text{th percentile}.
Dividing by the total number of scores (200200) and multiplying by 100100 converts the count into a percentage.

Anahtar Kavram

Percentile Rank Calculation
Tahmini Süre:1m 30s
Soru 2Soru

A data set consists of 6060 distinct test scores arranged in ascending order. The 70th70\text{th} percentile of this data set is equal to the 42nd42\text{nd} score. If 1515 new test scores, all strictly lower than the lowest score in the original data set, are added to form a new data set of 7575 scores, what is the percentile rank of the score that was the 70th70\text{th} percentile of the original data set?

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Cevap: 76

Cevap

76
The target score originally occupied position 42 in an ordered set of 60 values. When 15 values strictly smaller than all original values are introduced, they occupy the first 15 positions in the new ordered set. Consequently, the target score shifts to position 42 + 15 = 57 in the new set of 75 values. The percentile rank corresponds to the percentage of values at or below this position, calculated as (57 / 75) * 100% = 76%.

Adım Adım Çözüm

1
Determine the rank position of the original 70th percentile score
The target score is at position 42 in the original sorted set of 60 scores.
The stem specifies that the 70th percentile corresponds to the 42nd score.
2
Calculate the target score's position in the expanded data set
The target score is now at position 57 in the new sorted set of 75 scores.
Since 15 scores smaller than the original minimum are prepended, every original score's rank position increases by 15. Thus, position 42 becomes position 42 + 15 = 57.
3
Compute the percentile rank of position 57 out of 75
76
The percentile rank is the percentage of values less than or equal to this score in the new set: (57 / 75) * 100% = 76%.

Anahtar Kavram

Percentile Rank and Rank Position Shifts
Soru 3Soru

A quality control analyst measured the shelf life, in days, of a sample of 8080 manufactured batteries, where each battery had a distinct shelf life. Battery XX had a shelf life at the 65th65\text{th} percentile of the sample. If 2020 additional batteries are subsequently tested and every one of these 2020 batteries has a shelf life strictly greater than Battery XX, what is the percentile rank of Battery XX's shelf life in the combined sample of 100100 batteries?

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Cevap: 52nd52\text{nd} percentile

Cevap

The 52nd52\text{nd} percentile
In the initial sample of 8080 batteries, a shelf life at the 65th65\text{th} percentile means that 65%65\% of 8080 batteries, or 5252 batteries, have a shelf life less than or equal to Battery XX. When 2020 new batteries are added—all with shelf lives strictly greater than Battery XX—the new total number of batteries is 100100, while the number of batteries with shelf life less than or equal to Battery XX remains 5252. Thus, Battery XX is at the 52100×100%=52nd\frac{52}{100} \times 100\% = 52\text{nd} percentile of the combined sample.

Adım Adım Çözüm

1
Determine the number of batteries in the original sample with a shelf life less than or equal to Battery X
Since 65%65\% of the original 8080 batteries are at or below Battery XX, 0.65×80=520.65 \times 80 = 52 batteries.
By definition, a percentile rank of 6565 in a dataset of size NN represents 65%65\% of the data points at or below that score.
2
Calculate the total number of batteries in the combined dataset
80+20=10080 + 20 = 100 batteries.
Adding 2020 new batteries increases the sample size from 8080 to 100100.
3
Determine the number of batteries at or below Battery X in the combined dataset
The count remains 5252 batteries.
All 2020 newly added batteries have shelf lives strictly greater than Battery XX, so none of them fall at or below Battery XX.
4
Calculate the new percentile rank of Battery X
\frac{52}{100} \times 100\% = 52\text{nd percentile}.
The percentile rank is the percentage of total data points in the combined sample that are less than or equal to Battery XX.

Anahtar Kavram

Percentile rank represents the proportion of values in a dataset that are less than or equal to a given value. When a dataset expands by adding values strictly above a target value, the number of values at or below the target remains constant while the total sample size increases.
Tahmini Süre:1m 30s
Soru 4Soru

In a dataset of NN distinct test scores, score S1S_1 is at the 75th percentile and score S2S_2 is at the 40th percentile. A group of 60 new distinct scores is added to the dataset: 12 are strictly less than S2S_2, 18 are strictly between S2S_2 and S1S_1, and 30 are strictly greater than S1S_1. If score S1S_1 is at the 60th percentile of the combined dataset, what is the percentile rank of score S2S_2 in the combined dataset?

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Cevap: 28th percentile

Cevap

28th percentile
In the original dataset of NN scores, 0.75N0.75N scores are below S1S_1 and 0.40N0.40N scores are below S2S_2. Adding 60 scores (12 below S2S_2, 18 between S2S_2 and S1S_1, and 30 above S1S_1) increases the total dataset size to N+60N + 60. The number of scores below S1S_1 becomes 0.75N+12+18=0.75N+300.75N + 12 + 18 = 0.75N + 30. Given that S1S_1 is at the 60th percentile of the new dataset, 0.75N+30=0.60(N+60)0.75N + 30 = 0.60(N + 60), which simplifies to 0.15N=60.15N = 6, giving N=40N = 40. The original number of scores below S2S_2 is 0.40×40=160.40 \times 40 = 16. Adding the 12 new scores that are below S2S_2 gives 16+12=2816 + 12 = 28 scores below S2S_2 in the combined dataset. Out of 100 total scores in the combined dataset, the percentile rank of S2S_2 is 28100×100%=28%\frac{28}{100} \times 100\% = 28\%, which corresponds to the 28th percentile.

Adım Adım Çözüm

1
Set up expressions for the number of scores below S1S_1 and S2S_2 in the original dataset.
In the original dataset of NN scores, 0.75N0.75N scores are strictly less than S1S_1, and 0.40N0.40N scores are strictly less than S2S_2.
By definition of percentile rank, k%k\% percentile means k%k\% of the dataset scores fall strictly below that value.
2
Determine the number of scores strictly below S1S_1 in the combined dataset.
The total number of new scores added strictly below S1S_1 is 12+18=3012 + 18 = 30. Thus, the total number of scores below S1S_1 in the combined dataset is 0.75N+300.75N + 30, while the new total dataset size is N+60N + 60.
Scores added below S2S_2 and scores added between S2S_2 and S1S_1 are all strictly less than S1S_1.
3
Solve for NN using the 60th percentile rank condition for S1S_1 in the combined dataset.
0.75N+30=0.60(N+60)    0.75N+30=0.60N+36    0.15N=6    N=400.75N + 30 = 0.60(N + 60) \implies 0.75N + 30 = 0.60N + 36 \implies 0.15N = 6 \implies N = 40.
Setting the count of scores below S1S_1 equal to 60%60\% of the new total dataset size N+60N + 60 forms a single-variable linear equation.
4
Calculate the percentile rank of score S2S_2 in the combined dataset.
Original scores below S2=0.40×40=16S_2 = 0.40 \times 40 = 16. Combined scores below S2=16+12=28S_2 = 16 + 12 = 28. Combined total dataset size = 40+60=10040 + 60 = 100. Percentile rank of S2=28100×100%=28%S_2 = \frac{28}{100} \times 100\% = 28\%.
Dividing the total count of scores strictly below S2S_2 in the combined set by the total combined dataset size yields the updated percentile rank.

Anahtar Kavram

Percentile Rank and Combined Sets
Soru 5Soru

In a class of 4040 students, each student received a distinct score on a final examination. If a particular student's score is at the 85th85\text{th} percentile of the class scores, how many students scored lower than this student?

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Cevap: 3434

Cevap

34 students scored lower than this student.
The 85th percentile of a group of 40 students means that 85% of the class scored below this student. Calculating 85%85\% of 4040 yields 0.85×40=340.85 \times 40 = 34. Therefore, exactly 34 students scored lower.

Adım Adım Çözüm

1
Understand the definition of percentile rank.
Being at the 85th85\text{th} percentile means that the student's score is strictly higher than 85%85\% of the other scores in the dataset.
By definition, the pthp\text{th} percentile of a set of distinct values separates the lower p%p\% of the data from the rest.
2
Calculate 85%85\% of the total number of students (4040).
0.85×40=340.85 \times 40 = 34.
Multiplying the percentage by the sample size gives the exact count of students below the percentile threshold.

Anahtar Kavram

Percentile Rank
Soru 6Soru

A dataset consists of 20 distinct numerical values arranged in increasing order. The lower quartile (Q1Q_1) of the dataset is equal to the average of the 5th and 6th values, and the upper quartile (Q3Q_3) is equal to the average of the 15th and 16th values. Given that the lower quartile Q1=42Q_1 = 42, the interquartile range (IQR=Q3Q1\text{IQR} = Q_3 - Q_1) is 38, and the 15th value in the dataset is 74, what is the value of the 16th term?

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Cevap: 86

Cevap

86
To find the 16th value, first determine the upper quartile Q3Q_3 using the interquartile range: Q3=Q1+IQR=42+38=80Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80. Since Q3Q_3 is the average of the 15th and 16th terms, set up the equation 74+x162=80\frac{74 + x_{16}}{2} = 80. Multiplying by 2 gives 74+x16=16074 + x_{16} = 160, so x16=86x_{16} = 86.

Adım Adım Çözüm

1
Calculate the upper quartile (Q3Q_3) using the given lower quartile (Q1Q_1) and interquartile range (IQR).
Q3=Q1+IQR=42+38=80Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80
By definition, the interquartile range is the difference between the upper and lower quartiles (IQR=Q3Q1\text{IQR} = Q_3 - Q_1).
2
Relate the upper quartile (Q3Q_3) to the 15th and 16th values of the ordered dataset.
Q3=x15+x162=80Q_3 = \frac{x_{15} + x_{16}}{2} = 80
For an ordered dataset of 20 elements, Q3Q_3 is the median of the upper half of the data (the 11th through 20th terms), which equals the arithmetic mean of the 15th and 16th terms.
3
Substitute the known value of the 15th term (x15=74x_{15} = 74) into the equation and solve for the 16th term (x16x_{16}).
\frac{74 + x_{16}}{2} = 80 \implies 74 + x_{16} = 160 \implies x_{16} = 86
Multiplying both sides by 2 gives 160, and subtracting 74 yields 86.

Anahtar Kavram

Interquartile Range and Quartile Calculations
Soru 7Soru

In a corporate firm of NN employees, each employee earns a distinct annual salary. An employee earning a salary of $78,000\$78,000 is at the 60th60\text{th} percentile of all salaries in the firm. The firm subsequently hires 1010 new employees, each of whom earns an annual salary strictly less than $78,000\$78,000. If $78,000\$78,000 is now at the 70th70\text{th} percentile of all salaries in the expanded firm, how many employees were originally in the firm?

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Cevap: 30

Cevap

30
The percentile rank of a score indicates the percentage of values in the set that are less than or equal to that score. Originally, 60%60\% of NN employees earned $78,000\le \$78,000, giving 0.60N0.60N employees. Adding 1010 employees who all earn under $78,000\$78,000 brings the count of employees earning $78,000\le \$78,000 to 0.60N+100.60N + 10, while the total workforce becomes N+10N + 10. Since $78,000\$78,000 is at the 70th percentile of the new distribution, 0.60N+10=0.70(N+10)0.60N + 10 = 0.70(N + 10). Expanding and solving yields 0.60N+10=0.70N+70.60N + 10 = 0.70N + 7, so 0.10N=30.10N = 3, which gives N=30N = 30.

Adım Adım Çözüm

1
Define the initial number of employees earning at or below $78,000.
Initially, 0.60N0.60N employees earn $78,000\le \$78,000.
By definition of percentile rank with distinct values, being at the 60th percentile of NN scores means 60%60\% of the total NN salaries are less than or equal to $78,000\$78,000.
2
Determine the new count of employees and the updated count of employees earning at or below $78,000.
New total employees =N+10= N + 10; new count earning $78,000\le \$78,000 is 0.60N+100.60N + 10.
All 10 newly hired employees earn salaries strictly less than $78,000\$78,000, increasing the count of salaries $78,000\le \$78,000 by 10.
3
Set up an equation using the new percentile rank.
0.60N+10=0.70(N+10)0.60N + 10 = 0.70(N + 10)
The salary $78,000\$78,000 is now at the 70th percentile of the updated total group size of (N+10)(N + 10).
4
Solve the algebraic equation for NN.
0.60N+10=0.70N+7    3=0.10N    N=300.60N + 10 = 0.70N + 7 \implies 3 = 0.10N \implies N = 30.
Subtracting 0.60N0.60N and 77 from both sides gives 0.10N=30.10N = 3, which yields N=30N = 30.

Anahtar Kavram

Percentile Rank in Expanding Data Sets
Soru 8Soru

A dataset contains 2020 distinct test scores arranged in ascending order:

10,12,15,17,20,22,25,28,30,32,35,38,40,43,45,48,50,52,55,6010, 12, 15, 17, 20, 22, 25, 28, 30, 32, 35, 38, 40, 43, 45, 48, 50, 52, 55, 60

If the pthp\text{th} percentile of a dataset of NN values is defined as the value at position k=p100×Nk = \frac{p}{100} \times N when ordered from least to greatest, what is the 75th75\text{th} percentile of these test scores?

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Cevap: 45

Cevap

The 75th percentile of the given test scores is 45.
The 75th percentile corresponds to position k = (75/100) * 20 = 15 in the ordered list of 20 scores, which is equal to 45.

Adım Adım Çözüm

1
Calculate the position index kk for the 75th75\text{th} percentile.
k=75100×20=15k = \frac{75}{100} \times 20 = 15
The rank formula determines the 1-based index of the target percentile value in an ordered set of size N=20N = 20.
2
Locate the 15th15\text{th} score in the ordered dataset.
The 15th15\text{th} score is 4545.
Counting from the lowest score (1010 at position 1), the 15th15\text{th} score in the sequence is 4545.

Anahtar Kavram

Percentiles and Quartiles
Soru 9Soru

The table below shows the distribution of scores achieved by 5050 candidates on a professional certification assessment:

ScoreNumber of Candidates
606044
70701111
80802020
90901010
10010055

What is the interquartile range (IQR) of the assessment scores?

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Cevap: 2020

Cevap

The interquartile range of the assessment scores is 2020.
To find the interquartile range (IQR), first find Q1Q_1 (25th25\text{th} percentile) and Q3Q_3 (75th75\text{th} percentile). With 5050 total candidates, Q1Q_1 is situated around position 12.512.5, which falls into the score of 7070. Q3Q_3 is situated around position 37.537.5, which falls into the score of 9090. Subtracting Q1Q_1 from Q3Q_3 yields 9070=2090 - 70 = 20.

Adım Adım Çözüm

1
Calculate the cumulative frequency distribution to find score positions.
Score 6060: candidates 11 to 44; Score 7070: candidates 55 to 1515; Score 8080: candidates 1616 to 3535; Score 9090: candidates 3636 to 4545; Score 100100: candidates 4646 to 5050. Total N=50N = 50.
Cumulative frequencies identify the precise position of ranked scores.
2
Determine the first quartile (Q1Q_1), which represents the 25th25\text{th} percentile.
The 25th25\text{th} percentile corresponds to the 0.25×50=12.5th0.25 \times 50 = 12.5\text{th} position. Looking at the cumulative frequencies, candidate positions 55 through 1515 all scored 7070, so Q1=70Q_1 = 70.
The first quartile marks the score boundary below which 25%25\% of the dataset falls.
3
Determine the third quartile (Q3Q_3), which represents the 75th75\text{th} percentile.
The 75th75\text{th} percentile corresponds to the 0.75×50=37.5th0.75 \times 50 = 37.5\text{th} position. Candidate positions 3636 through 4545 all scored 9090, so Q3=90Q_3 = 90.
The third quartile marks the score boundary below which 75%75\% of the dataset falls.
4
Compute the interquartile range (IQR=Q3Q1)(\text{IQR} = Q_3 - Q_1).
\text{IQR} = 90 - 70 = 20.
The interquartile range measures the spread of the middle 50%50\% of the distribution.

Anahtar Kavram

Interquartile Range (IQR) and Quartile Positions in Frequency Distributions
Soru 10Soru

In a dataset of NN distinct values arranged in ascending order, the pthp\text{th} percentile is defined as the value at position p100×N\left\lceil \frac{p}{100} \times N \right\rceil.

Initially, a dataset contains 100100 distinct test scores. The 40th40\text{th} percentile score is S1S_1 and the 70th70\text{th} percentile score is S2S_2. A researcher adds xx new scores that are strictly less than S1S_1, and yy new scores that are strictly between S1S_1 and S2S_2, where xx and yy are positive integers.

If the 40th40\text{th} percentile score of the expanded dataset remains S1S_1 and the 70th70\text{th} percentile score of the expanded dataset remains S2S_2, what is the maximum possible value of x+yx + y?

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Cevap: 33

Cevap

The maximum possible value of x+yx + y is 33.
The correct choice is 33. By setting up the percentile rank equations using the ceiling function, the position of S1S_1 in the expanded dataset requires 0.4(x+y)=x\lceil 0.4(x + y) \rceil = x, while the position of S2S_2 requires 0.7(x+y)=x+y\lceil 0.7(x + y) \rceil = x + y. Letting k=x+yk = x + y, the ceiling condition 0.7k=k\lceil 0.7k \rceil = k implies k1<0.7kkk - 1 < 0.7k \le k, which simplifies to 0.3k<10.3k < 1 or k<3.33k < 3.33. Since xx and yy are positive integers, kk can take integer values of 22 or 33. Thus, the maximum possible value of x+yx + y is 33.

Adım Adım Çözüm

1
Determine the initial rank positions of S1S_1 and S2S_2 in the 100-element dataset.
Position of S1=0.40×100=40S_1 = \lceil 0.40 \times 100 \rceil = 40; Position of S2=0.70×100=70S_2 = \lceil 0.70 \times 100 \rceil = 70.
Applying the given percentile formula to N=100N = 100 gives the exact index of each score in ascending order.
2
Express the new rank positions of S1S_1 and S2S_2 after inserting xx scores below S1S_1 and yy scores between S1S_1 and S2S_2.
In the new dataset of size 100+x+y100 + x + y, S1S_1 is at position 40+x40 + x and S2S_2 is at position 70+x+y70 + x + y.
Adding xx scores less than S1S_1 shifts S1S_1's rank by xx. Adding yy scores between S1S_1 and S2S_2 further shifts S2S_2's rank by x+yx + y relative to its original position.
3
Formulate the percentile conditions for the expanded dataset of size M=100+x+yM = 100 + x + y.
For S1S_1: 0.40(100+x+y)=40+x    0.4(x+y)=x\lceil 0.40(100 + x + y) \rceil = 40 + x \implies \lceil 0.4(x + y) \rceil = x.
For S2S_2: 0.70(100+x+y)=70+x+y    0.7(x+y)=x+y\lceil 0.70(100 + x + y) \rceil = 70 + x + y \implies \lceil 0.7(x + y) \rceil = x + y.
The percentile rank of S1S_1 must equal its new position 40+x40 + x, and the percentile rank of S2S_2 must equal its new position 70+x+y70 + x + y.
4
Solve for the total added elements k=x+yk = x + y.
From 0.7k=k\lceil 0.7k \rceil = k, we get k1<0.7kkk - 1 < 0.7k \le k, which simplifies to 0.3k<1    k<1033.330.3k < 1 \implies k < \frac{10}{3} \approx 3.33.
By definition of the ceiling function, z=k    k1<zk\lceil z \rceil = k \iff k - 1 < z \le k for integer kk.
5
Evaluate integer solutions for k=x+yk = x + y given x,y1x, y \ge 1.
For k=2k = 2: x=0.4(2)=1x = \lceil 0.4(2) \rceil = 1, y=1y = 1 (valid).
For k=3k = 3: x=0.4(3)=2x = \lceil 0.4(3) \rceil = 2, y=1y = 1 (valid).
The maximum integer value for k=x+yk = x + y is 33.
Since xx and yy are positive integers, k2k \ge 2. The maximum integer k<3.33k < 3.33 is 33.

Anahtar Kavram

Percentile Rank and Indexing with Ceiling Functions
Tahmini Süre:2m 30s