Home / Aptitude Tests / Mathematics / \(\begin{array}{c|c}...
\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 &...

\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & y + 2 & y - 2 & 2y - 3 & y + 4 & 3y - 4\end{array}\)
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y

A.

1

B.

2

C.

3

D.

4

Correct answer is B

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 = \(\frac{\sum fx}{\sum f}\)

\(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\)

\(\implies 14(28y - 13) = 43(8y - 2)\)

\(392y - 182 = 344y - 86\)

\(392y - 344y = -86 + 182 \implies 48y = 96\)

\(y = 2\)