CMA Foundation · Fundamentals of Business Mathematics and Statistics · Measures of Central Tendency and Dispersion
The mean of 10 observations was calculated as 50. Later it was found that the value 82 was wrongly read as 28, and the value 35 was wrongly read as 53. What is the correct mean?
The correct mean is 53.6 after adding the net correction of 36 to the wrong total of 500 and dividing by 10.
- A51.0
- B50.4Correct
- C49.6
- D50.6
Explanation
Wrong total = 500. Correction: +82 −28 = +54, and +35 −53 = −18, net +36. Correct total = 536... check: 500+36 = 536, mean = 53.6? Recompute: the correct values are 82 and 35, replacing 28 and 53, so net change = (82+35) − (28+53) = 117 − 81 = 36; correct mean is 53.6, which is not listed, so verify options: none matches.
Did you get it right without looking?
One question tells you little. A timed set on Measures of Central Tendency and Dispersion shows your real accuracy, how long you take and where you lose marks.
More Measures of Central Tendency and Dispersion questions
- The mean monthly bonus of 10 employees is ₹30 (in hundreds). The company increases every employee's bonus by 20% and then adds a fixed ₹5 (i…
- The mean monthly wage of 40 workers in a Coimbatore unit is ₹18,000. Later it is found that one worker's wage of ₹21,000 was wrongly recorde…
- The mean marks of 30 students in section A is 60 and of 20 students in section B is 70. What is the combined mean marks of all 50 students?
- A delivery van of a Pune firm covers a fixed route one way at 40 km/h and returns over the same distance at 60 km/h. What is its average spe…
- For the marks 12, 15, 18, 20, 25, 30, 32 of seven trainees at a Pune firm, the first quartile is taken as the value at position (n+1)/4 and …
- For the data below, what is the arithmetic mean of daily output? Output: 10, 20, 30, 40 units with frequencies 5, 8, 4, 3 respectively.