Teachers Work Experience Statistics

XAT 2024 · Quantitative Ability & Data Interpretation · Hard · Statistics

Passage / data set

The upper hinge of a dataset is the median of all the values to the right of the median of the dataset in an ascending arrangement, while the lower hinge of the dataset is the median of all the values to the left of the median of the dataset in the same arrangement.

For example, consider the dataset 4, 3, 2, 6, 4, 2, 7. When arranged in ascending order, it becomes 2, 2, 3, 4, 4, 6, 7. The median is 4 (the bold value), and hence the upper hinge is the median of 4, 6, 7, i.e., 6. Similarly, the lower hinge is 2.

A student has surveyed thirteen of her teachers, and recorded their work experience (in integer years). Two of the values recorded by the student got smudged, and she cannot recall those values. All she remembers is that those two values were unequal, so let us write them as AA and BB, where A<BA < B. The remaining eleven values, as recorded, are: 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29. Moreover, the student also remembers the following summary measures, calculated based on all the thirteen values:

  • Minimum: 2
  • Lower Hinge: 6.5
  • Median: 12
  • Upper Hinge: 21
  • Maximum: 29

Question 1 of 3

Which of the following is a possible value of BB?

  1. A.

    2

  2. B.

    6

  3. C.

    8

  4. D.

    13

  5. E.

    29

Answer

C

Explanation

Since the minimum value is 2 and 2 is not among the 11 known values, the smaller smudged value AA must be 2. With A=2A = 2, we have 12 values: 2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29 plus BB. The median of 13 values is the 7th7^{\text{th}} value in sorted order, which is given as 12. For 12 to be the 7th7^{\text{th}} element, BB must be ≤12\le 12. The lower hinge is the median of the first 6 values. The first 6 values are 2, 5, 6, 7, 8, and BB. For their median to be 6.5, BB must be greater than 7 (so the middle two values are 6 and BB, giving (6+B)/2=6.5  ⟹  B=7(6+B)/2 = 6.5 \implies B = 7, or BB can be 8, 9, 10, 11, 12 giving lower hinge between 6.5 and 7). Among the options, 8 lies in the valid range 7≤B≤127 \le B \le 12.

Question 2 of 3

Based on the information recorded, which of the following can be the average work experience of the thirteen teachers?

  1. A.

    12

  2. B.

    12.5

  3. C.

    13

  4. D.

    13.5

  5. E.

    14

Answer

E

Explanation

Sum of 11 known values =5+6+7+8+12+16+19+21+21+27+29=171= 5+6+7+8+12+16+19+21+21+27+29 = 171. Adding A=2A = 2, sum of 12 values =173= 173. BB can range from 7 to 12. If B=9B = 9, the total sum =173+9=182= 173 + 9 = 182. Average =182/13=14= 182 / 13 = 14.

Question 3 of 3

While rechecking her original notes to re-enter the smudged values of AA and BB in the records, the student found that one of the eleven recorded work experience values that did not get smudged was recorded wrongly as half of its correct value. After re-entering the values of AA and BB, and correcting the wrongly recorded value, she recalculated all the summary measures. The recalculated average value was 15. What is the value of BB?

  1. A.

    7

  2. B.

    9

  3. C.

    10

  4. D.

    12

  5. E.

    Cannot be determined from the given information.

Answer

C

Explanation

Recalculated average =15  ⟹  = 15 \implies Recalculated total sum =15×13=195= 15 \times 13 = 195. The sum before correction was 173+B173 + B. Correcting a value that was recorded as half means adding its original value UU to the sum. 173+B+U=195  ⟹  B+U=22173 + B + U = 195 \implies B + U = 22. B∈{7,8,9,10,11,12}B \in \{7, 8, 9, 10, 11, 12\} and U∈{5,6,7,8,12,16,19,21,27,29}U \in \{5, 6, 7, 8, 12, 16, 19, 21, 27, 29\}. Testing pairs where B+U=22B + U = 22:

  • If B=10B = 10, U=12U = 12, which is a valid value from the list. Thus B=10B = 10.

Related Statistics questions

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace