Descriptive Statistics
A group of ten numbers has a recorded mean of fifteen, but later one entry originally read as eight is found to be twenty-three; what is the corrected mean of the ten numbers?
Select the correct option:
Solution
16.5
The mean equals the total sum divided by the number of observations, so any correction to a single value changes the sum by the difference and shifts the mean by that difference spread over all entries. This sum-adjustment idea is a routine JEE Advanced descriptive-statistics manoeuvre. The original total of ten numbers is mean × count = 15 × 10 = 150. The wrong entry 8 must be replaced by the correct value 23, increasing the sum by 23 − 8 = 15, giving a corrected total of 150 + 15 = 165. Dividing by 10 yields the corrected mean 165/10 = 16.5. Option 16.0 results from adding only 10 instead of the true difference 15. Option 15.5 comes from spreading half the correction. Option 17.0 over-adds by using a difference of 20. The principle is that the mean responds linearly to the corrected sum, here total/count. Plausibility check: replacing 8 by the larger 23 must raise the mean, and the increase of 1.5 equals the correction 15 divided by the 10 observations, exactly as the averaging mechanism predicts.
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About This Question
- Subject
- mathematics
- Chapter
- statistics and probability
- Topic
- descriptive statistics
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
16.5
The mean equals the total sum divided by the number of observations, so any correction to a single value changes the sum by the difference and shifts the mean by that difference spread over all entries. This sum-adjustment idea is a routine JEE Advanced descriptive-statistics manoeuvre. The original total of ten numbers is mean × count = 15 × 10 = 150. The wrong entry 8 must be replaced by the correct value 23, increasing the sum by 23 − 8 = 15, giving a corrected total of 150 + 15 = 165. Dividing by 10 yields the corrected mean 165/10 = 16.5. Option 16.0 results from adding only 10 instead of the true difference 15. Option 15.5 comes from spreading half the correction. Option 17.0 over-adds by using a difference of 20. The principle is that the mean responds linearly to the corrected sum, here total/count. Plausibility check: replacing 8 by the larger 23 must raise the mean, and the increase of 1.5 equals the correction 15 divided by the 10 observations, exactly as the averaging mechanism predicts.
This easy difficulty mathematics question is from the chapter statistics and probability, covering the topic of descriptive statistics. It appeared in the 2025 exam.
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