If the mean of the data Class, 5--10, 10--15, 15--20, 20--25, 25--30, 30--35 | ---|---|---…

Question

If the mean of the data
 Class, 5--10, 10--15, 15--20, 20--25, 25--30, 30--35 |
---|---|---|---|---|---|---|
 Frequency, 2, $k$, 28, 54, $k+1$, 5 |

is 21, then $k$ is one of the roots of the equation:

Accepted Answer

Class marks: $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$

$$\sum f_i = 2+k+28+54+(k+1)+5 = 2k+90$$

$$\sum f_ix_i = 2(7.5)+k(12.5)+28(17.5)+54(22.5)+(k+1)(27.5)+5(32.5)$$
$$= 15+12.5k+490+1215+27.5k+27.5+162.5 = 40k+1910$$

Mean: $\dfrac{40k+1910}{2k+90} = 21$

$$21(2k+90) = 40k+1910 \implies 42k+1890 = 40k+1910 \implies 2k = 20 \implies k=10$$

Checking $k=10$ in each option, $2x^2-19x-10=0$ gives $2(100)-19(10)-10 = 200-190-10=0$.  So Option C is satisfied by $k=10$.

Solutions · 1