Question

- The mean age of a group of 40 students is 17.45 years. Find the missing frequencies.
Age (in years) | 15 16 17 18 19 20
Number of students 3 ? 9 11 ? 3
3
​

Answer 1

Answer:

Missing Frequencies are 8 and 6 respectively

Step-by-step explanation:

We know that mean is calculated by the formula given

$Mean = \frac{ \Sigma \: x_if_i}{ \Sigma \: f_i} \\ \\ = \frac{sum \: of \: all \: observations}{total \: number \: of \: observation}$

So

$17.45 = \frac{15 \times 3 + 16x + 17 \times 9 + 18 \times 11 + 19y + 20 \times 3}{40} \\ \\ 698 = 45 + 16x + 153 + 198 + 19y + 60 \\ \\ 698 = 456 + 16x + 19y \\ \\ 242 = 16x + 19y \\ \\ or \\ \\ 16x + 19y = 242 \: ....eq1$

As the total number of students are 40,so

$3 + x + 9 + 11 + y + 3 = 40 \\ \\ x + y = 14 \: .....eq2$

Now calculate the values of x and y from eq1 and eq2

$16x + 19y = 242 \\ 16x + 16y = 224 \\ ( - ) \: \: \: \: ( - ) \: \: \: \: \: ( - ) \\ - - - - - - - - - \\ 3y = 18 \\ \\ y = 6$

So , now put the value of y in eq2

$x + 6 = 14 \\ \\ x = 14 - 6 \\ \\ x = 8$

So, the missing Frequencies are 6 and 8.

Hope it helps you.

Answer 2

Step-by-step explanation:

mark as brilliant

hope you like it

thank you