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 = \(\frac{\sum f(x - x)^2}{\sum f}\)
= \(\frac{90}{18}\)
= \(\sqrt{5}\)
25
15
10
8
Correct answer is C
Total number of students (y) using the subject French
( 90 / 360 )º x Y = 30
Y = 30 x 4
Y = 120
History- 90º + French- 90º + Economics- 150º + CRK = 360º
CRK = 360º - 330º = 30º
Number that offered CRK:
( 30 / 360 )º x 120 = 10
27o
30o
54o
108o
Correct answer is D
Number of 15- year old pupils = 30.
Total number of students: 3 + 10 + 30 + 42 + 15 = 100.
Angle = \(\frac{30}{100} \times 360°\)
= \(108°\)
24
22
21
20
Correct answer is D
\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x + 1\)
\(y = \int_{1} ^{3} (3x^{2} - 2x + 1) \mathrm d x\)
\(y = [x^{3} - x^{2} + x]_{1} ^{3}\)
= \([3^{3} - 3^{2} + 3] - [1^{3} - 1^{2} + 1]\)
= \(21 - 1 = 20\)
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