\(\frac{7}{3}\)
\(\frac{2}{3}\)
\(\frac{1}{3}\)
\(\frac{3}{10}\)
Correct answer is D
No explanation has been provided for this answer.
4.0
3.4
3.2
3.0
Correct answer is B
Median = \(\frac{a + b}{fm}\) (\(\frac{1}{2} \sum f - CF_b\))
= 2.95 + \(\frac{0.5}{15}\)(2.-7)
= 2.95 + \(\frac{0.5}{15}\) x 13
= 2.95 + 0. 43
= 3.38
= 3.4
3.2
3.3
3.7
4.2
Correct answer is B
Mode = a + (b - a)(fm - Fb)
2Fm - Fa - Fb
= 3.0 + \(\frac{(3.4 - 3)(15 - 4)}{2(15) - 4 - 10}\)
= 3 + \(\frac{(6.4)(11)}{30 - 14}\)
= 3 + \(\frac{4.4}{16}\)
= 3 + 0.275
= 3.275
= 3.3cm
The variance of the scores 1, 2, 3, 4, 5 is
1, 2
1, 4
2.0
3.0
Correct answer is C
| \(x\) | 1 | 2 | 3 | 4 | 5 | sum =15 |
| \(x - \bar{x}\) | -2 | -1 | 0 | 1 | 2 | |
| \((x - \bar{x})^{2}\) | 4 | 1 | 0 | 1 | 4 | 10 |
Mean = \(\frac{15}{5} = 3\)
\(Variance = \frac{\sum (x - \bar{x})^{2}}{n}\)
= \(\frac{10}{5}\)
= \(2.0\)
5
\(\sqrt{6}\)
\(\frac{5}{3}\)
\(\sqrt{5}\)
Correct answer is D
\(\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 2 & 10 & -3 & 9 & 45\\ 4 - 6 & 8 & 5 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 8 & 40 & 3 & 9 & 45 \\ \hline & 18 & & 90 & & & 90 \end{array}\)
x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{90}{18}\)
= 5
S.D = \(\frac{\sum f(x - x)^2}{\sum f}\)
= \(\frac{90}{18}\)
= \(\sqrt{5}\)