Let the mean and the variance of 5 observations x1, x2, x3, x4, x5 be \(\frac{24}{5} \) and \(\frac{194}{25}\), respectively. If the mean and variance of the first 4 observations are \(\frac{7}{2}\) and a, respectively, then (4a + x5) is equal to
- 13
- 15
- 17
- 18
The Correct Option is B
Solution and Explanation
The correct option is(B): 15
\(\sum_i=1^5x_i=24....(i)\)
\(\sum{x^2_i=1154}....(i)\)
⇒ x5 = 10
⇒ x5 + 4a = 10 + 5 = 15
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JEE Main Notification
GMC Azamgarh Admission 2025: Dates, Courses, Fees, Eligibility, SelectJuly 22, 2025Concepts Used:
Variance and Standard Deviation
Variance:
According to layman’s words, the variance is a measure of how far a set of data are dispersed out from their mean or average value. It is denoted as ‘σ2’.
Variance Formula:
Read More: Difference Between Variance and Standard Deviation
Standard Deviation:
The spread of statistical data is measured by the standard deviation. Distribution measures the deviation of data from its mean or average position. The degree of dispersion is computed by the method of estimating the deviation of data points. It is denoted by the symbol, ‘σ’.
Types of Standard Deviation:
- Standard Deviation for Discrete Frequency distribution
- Standard Deviation for Continuous Frequency distribution
Standard Deviation Formulas:
1. Population Standard Deviation
2. Sample Standard Deviation
