\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & y + 2 & y – 2 & 2y – 3 & y + 4 & 3y – 4\end{array}\)
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y
- 1
-
2
- 3
- 4
The correct answer is: B
Explanation
| x | 1 | 2 | 3 | 4 | 5 | Total |
| f | y + 2 | y - 1 | 2y - 3 | y + 4 | 3y - 4 | 8y - 2 |
| fx | y + 2 | 2y - 2 | 6y - 9 | 4y + 16 | 15y - 20 | 28y - 13 |
Mean = \(\frac{\sum fx}{\sum f}\)
\(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\)
\(\implies 14(28y - 13) = 43(8y - 2)\)
\(392y - 182 = 344y - 86\)
\(392y - 344y = -86 + 182 \implies 48y = 96\)
\(y = 2\)