If the mean of the data Class5-1010-1515-2020-2525-3030-35Frequency2k2854k + 15is 21, then…

Question

If the mean of the data
Class5-1010-1515-2020-2525-3030-35Frequency2k2854k + 15is 21, then $k$ is one of the roots of the equation:

Accepted Answer

For the grouped data, we first find the class marks of each class interval.

The classes are:

$5-10,\, 10-15,\, 15-20,\, 20-25,\, 25-30,\, 30-35$

So their class marks are:

$7.5,\, 12.5,\, 17.5,\, 22.5,\, 27.5,\, 32.5$

The frequencies are:

$2,\, k,\, 28,\, 54,\, k+1,\, 5$

Now use the formula for mean of grouped data:

$\vec{x}=\frac{\sum {f}_{i}{x}_{i}}{\sum {f}_{i}}$

Given mean $=21$.

Now calculate $\sum {f}_{i}$:

$\sum {f}_{i}=2+k+28+54+(k+1)+5$

$\sum {f}_{i}=2k+90$

Now calculate $\sum {f}_{i}{x}_{i}$:

$\sum {f}_{i}{x}_{i}=2(7.5)+k(12.5)+28(17.5)+54(22.5)+(k+1)(27.5)+5(32.5)$

Now simplify step by step:

$2(7.5)=15$

$28(17.5)=490$

$54(22.5)=1215$

$5(32.5)=162.5$

$(k+1)(27.5)=27.5k+27.5$

So,

$\sum {f}_{i}{x}_{i}=15+12.5k+490+1215+27.5k+27.5+162.5$

$\sum {f}_{i}{x}_{i}=40k+1910$

Now apply the mean formula:

$21=\frac{40k+1910}{2k+90}$

Cross-multiply:

$21(2k+90)=40k+1910$

$42k+1890=40k+1910$

$2k=20$

$k=10$

Now we check which equation has $k=10$ as a root.

Option A:

$2{x}^{2}-23x-10=0$

Put $x=10$:

$2(10{)}^{2}-23(10)-10=200-230-10=-40
e 0$

Option B:

$4{x}^{2}-35x+24=0$

Put $x=10$:

$4(10{)}^{2}-35(10)+24=400-350+24=74
e 0$

Option C:

$2{x}^{2}-19x-10=0$

Put $x=10$:

$2(10{)}^{2}-19(10)-10=200-190-10=0$

So $10$ is a root.

Hence, the correct option is:

$\, 2{x}^{2}-19x-10=0$

Solutions · 1