Question
Download Solution PDFIf the mean of a frequency distribution is 100 and the coefficient of variation is 45%, then what is the value of the variance?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)
Variance = (Standard Deviation)2
Calculation:
Given coefficient of variation = 45% = 0.45
And mean = 100
As Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)
0.45 = \(\rm\text{Standard Deviation} \over100\)
Standard Deviation = 100 × 0.45
SD = 45
∴ Variance = 452 = 2025
Last updated on Jun 18, 2025
->UPSC has extended the UPSC NDA 2 Registration Date till 20th June 2025.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.