Jambmaths question:
|
Value |
0 |
1 |
2 |
3 |
4 |
|
Frequency |
1 |
2 |
2 |
1 |
9 |
Find the mean of the distribution above
Option A:
1
Option B:
2
Option C:
3
Option D:
4
Jamb Maths Solution:
$\begin{align} & mean=\overline{x}=\frac{(0\times 1)+(1\times 2)+(2\times 2)+(3\times 1)+(4\times 9)}{1+2+2+1+9} \\ & \overline{x}=\frac{0+2+4+3+36}{15} \\ & \overline{x}=\frac{45}{15}=3 \end{align}$
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Year of Exam: