Calculate the average of the following discrete series. Use the short-cut method by taking 25 as the assumed average. Calculate average marks.

Question image

Understand the Problem

The image contains two separate math problems related to calculating averages from discrete series.

Problem 5 asks to calculate the average of a discrete series, and to use the short-cut method, assuming 25 as the average.

Problem 6 asks to find the average marks secured by 42 students in Economics, given the marks and the number of students who secured those marks.

Answer

5. $24.925$ 6. $23.19$
Answer for screen readers
  1. The average of the discrete series is $24.925$.
  2. The average marks secured by the students is approximately $23.19$.

Steps to Solve

  1. Calculate the deviations from the assumed mean for Problem 5

The assumed mean is 25. Calculate $d_i = x_i - A$, where $x_i$ is the size, and $A$ is the assumed mean (25). $d_1 = 30 - 25 = 5$ $d_2 = 29 - 25 = 4$ $d_3 = 28 - 25 = 3$ $d_4 = 27 - 25 = 2$ $d_5 = 26 - 25 = 1$ $d_6 = 25 - 25 = 0$ $d_7 = 24 - 25 = -1$ $d_8 = 23 - 25 = -2$ $d_9 = 22 - 25 = -3$ $d_{10} = 21 - 25 = -4$

  1. Calculate $f_i d_i$ for Problem 5

Multiply each deviation $d_i$ by its corresponding frequency $f_i$. $f_1 d_1 = 2 \times 5 = 10$ $f_2 d_2 = 4 \times 4 = 16$ $f_3 d_3 = 5 \times 3 = 15$ $f_4 d_4 = 3 \times 2 = 6$ $f_5 d_5 = 2 \times 1 = 2$ $f_6 d_6 = 7 \times 0 = 0$ $f_7 d_7 = 1 \times -1 = -1$ $f_8 d_8 = 4 \times -2 = -8$ $f_9 d_9 = 5 \times -3 = -15$ $f_{10} d_{10} = 7 \times -4 = -28$

  1. Calculate $\sum f_i d_i$ for Problem 5

Sum all the $f_i d_i$ values. $\sum f_i d_i = 10 + 16 + 15 + 6 + 2 + 0 - 1 - 8 - 15 - 28 = -3$

  1. Calculate $\sum f_i$ for Problem 5

Sum all the frequencies. $\sum f_i = 2 + 4 + 5 + 3 + 2 + 7 + 1 + 4 + 5 + 7 = 40$

  1. Apply the short-cut formula for Problem 5

The formula for the average using the short-cut method is: $\text{Average} = A + \frac{\sum f_i d_i}{\sum f_i}$, where $A$ is the assumed mean. $\text{Average} = 25 + \frac{-3}{40} = 25 - 0.075 = 24.925$

  1. Calculate $f_i x_i$ for Problem 6

Multiply each mark $x_i$ by its corresponding number of students $f_i$. $f_1 x_1 = 8 \times 15 = 120$ $f_2 x_2 = 4 \times 20 = 80$ $f_3 x_3 = 7 \times 22 = 154$ $f_4 x_4 = 3 \times 23 = 69$ $f_5 x_5 = 8 \times 27 = 216$ $f_6 x_6 = 7 \times 35 = 245$ $f_7 x_7 = 5 \times 18 = 90$

  1. Calculate $\sum f_i x_i$ for Problem 6

Sum all the $f_i x_i$ values. $\sum f_i x_i = 120 + 80 + 154 + 69 + 216 + 245 + 90 = 974$

  1. Calculate the average for Problem 6

The average is $\frac{\sum f_i x_i}{\sum f_i}$. We are given that the total number of students ($\sum f_i$) is 42. $\text{Average} = \frac{974}{42} \approx 23.19$

  1. The average of the discrete series is $24.925$.
  2. The average marks secured by the students is approximately $23.19$.

More Information

The short-cut method is a way to simplify calculations of the mean, especially when dealing with larger numbers. It works by choosing an arbitrary value as the assumed mean, calculating deviations from this value, and then correcting for the deviations.

Tips

A common mistake is in calculating the deviations $d_i$ incorrectly, especially when the sizes are smaller than the assumed mean, in which can lead to errors in sign. Another common mistake is messing up the arithmetic.

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