Question

The following data give the percentage of criminals for different age groups:

Age (less than X) 25 30 40 50 55 65

% of criminals 52 67.3 84.1 94.4 104.7 M

In the above table the value of M depends on Student ID. Suppose, one student ID

is 1503001, then the value of M would be: 001+100 meaning that last 3digits of

student ID+100. If anyone ID is 1503101 then the value of M would be: 101+100. The

value of M should be considered last 3 digits of student ID+100 and proceed

successively.

Using Lagrange’s formula, find the percentage of criminals under the age of 45.

Answer 1

$f(x)= \frac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)}y_0+ \frac{(x-x_0)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)} y_1+ \frac{(x-x_0)(x-x_1)(x-x_3)(x-x_4)(x-x_5)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)(x_2-x_5)}y_2+ \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_4)(x-x_5)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)(x-x_4)(x-x_5)} y_3+ \frac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_5)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)(x_4-x_5)} y_4$

$f(x)= \frac{(45-30)(45-40)(45-50)(45-55)(45-65)}{(25-30)(25-40)(25-50)(25-55)(25-65)}52+ \frac{(45-25)(45-40)(45-50)(45-55)(45-65)}{(30-25)(30-40)(30-50)(30-55)(30-65)} 67.3+ \frac{(45-25)(45-30)(45-50)(45-55)(45-65)}{(40-25)(40-30)(40-50)(40-55)(40-65)}84.1+ \frac{(45-25)(45-30)(45-40)(45-55)(45-65)}{(50-25)(50-30)(50-40)(50-55)(50-65)} 94.4+ \frac{(45-25)(45-30)(45-40)(45-50)(45-65)}{(55-25)(55-30)(55-40)(55-50)(55-65)}104.7$

$=\frac{1}{30}\cdot52 -\frac{4}{35}\cdot67.3 +\frac{8}{15}\cdot84.1 + \frac{4}{5}\cdot94.4 -\frac{4}{15}\cdot104.7$

$=86.5\%$