JAMB - Mathematics (1992 - No. 38)
\(\begin{array}{c|c} x & 2 & 4 & 6 & 8\\ \hline f & 4 & y & 6 & 5 \end{array}\)
If the mean of the above frequency distribution is 5.2, find y
If the mean of the above frequency distribution is 5.2, find y
6.0
5.2
5.0
4.0
Explanation
Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{5.2}{1}\)
= \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\)
= \(\frac{5.2}{1}\) = \(\frac{84 + 4y}{15 + y}\)
= 5.2(15 + y) = 84 + 4y
= 84 + 4y = 78 + 5.2y
= 5.2y - 4y = 84 - 78
1.2y = 6
y = \(\frac{6}{1.2}\)
= \(\frac{60}{12}\)
= 5
= \(\frac{5.2}{1}\)
= \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\)
= \(\frac{5.2}{1}\) = \(\frac{84 + 4y}{15 + y}\)
= 5.2(15 + y) = 84 + 4y
= 84 + 4y = 78 + 5.2y
= 5.2y - 4y = 84 - 78
1.2y = 6
y = \(\frac{6}{1.2}\)
= \(\frac{60}{12}\)
= 5
Comments (0)
