If a variable takes the discrete values $\alpha + 4, \alpha - \frac{7}{2}, \alpha - \frac{5}{2}, \alpha - 3, \alpha - 2, \alpha + \frac{1}{2}, \alpha - \frac{1}{2}, \alpha + 5$ where $\alpha > 0$,then the median of these values is:
- A$\alpha - \frac{5}{4}$
- B$\alpha - \frac{1}{2}$
- C$\alpha - 2$
- D$\alpha + \frac{5}{4}$
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Consider a set of observations ${x_1}, {x_2}, {x_3}, ..., {x_{101}}$ such that ${x_1} < {x_2} < {x_3} < ... < {x_{100}} < {x_{101}}$. The mean deviation of this set of observations about a point $k$ is minimum when $k$ equals:
Easy
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View SolutionIf for some $x \in R$,the frequency distribution of the marks obtained by $20$ students in a test is:
Marks: $2, 3, 5, 7$
Frequency: $(x+1)^2, 2x-5, x^2-3x, x$
Then the mean of the marks is:
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Frequency: $(x+1)^2, 2x-5, x^2-3x, x$
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The mean marks of $100$ students in a class in mathematics is $72$. If the number of boys is $70$ and their mean marks is $75$,find the mean marks of the girls in the class.
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View SolutionWhich formula is used to calculate the weighted mean?
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