Ch 13: Statistics (Part 1)
Master the Direct, Assumed Mean, and Step-Deviation methods for Mean, and the formula for Mode.
Exercise 13.1 (Finding the Mean)
💡 Trick: Use the Direct Method (Σfi xi / Σfi) if values are small. Use Assumed Mean (a + Σfi di / Σfi) for large values.
Q1. A survey was conducted to find the mean number of plants per house. (0-2, 2-4, ... 12-14).
Solution (Direct Method):
Since data values are small, Direct Method is easiest.
$\Sigma f_i = 20$ (Total houses)
$\Sigma f_i x_i = 162$
Mean = $\Sigma f_i x_i / \Sigma f_i = 162 / 20 =$ 8.1 plants.
Q2. Find the mean daily wages of 50 workers of a factory. (500-520, 520-540...)
Solution (Assumed Mean Method):
Let assumed mean $a = 550$. Class size $h = 20$.
$\Sigma f_i = 50$
$\Sigma f_i d_i = -240$
Mean = $a + (\Sigma f_i d_i / \Sigma f_i) = 550 + (-240 / 50) = 550 - 4.8 =$ ₹ 545.20.
Q3. The following distribution shows the daily pocket allowance. The mean pocket allowance is ₹18. Find the missing frequency f.
Solution:
Given Mean = 18. Let's use Direct Method.
$\Sigma f_i = 44 + f$
$\Sigma f_i x_i = 752 + 20f$
Mean = $\Sigma f_i x_i / \Sigma f_i \implies 18 = (752 + 20f) / (44 + f)$
$18(44 + f) = 752 + 20f$
$792 + 18f = 752 + 20f$
$40 = 2f \implies$ $f = 20$.
Q4. Heartbeats per minute of 30 women. Find mean. (65-68, 68-71...)
Assumed Mean $a = 75.5$. $\Sigma f_i d_i = 12$. $\Sigma f_i = 30$.
Mean = $75.5 + (12/30) =$ 75.9 beats/min.
Q5. Mangoes in packing boxes. (50-52, 53-55...) Find mean.
Note: Data is discontinuous. Make it continuous (49.5-52.5, 52.5-55.5).
Assumed Mean $a = 57$. $\Sigma f_i d_i = 75$. $\Sigma f_i = 400$.
Mean = $57 + (75/400) =$ 57.19 mangoes.
Q6. Daily expenditure on food of 25 households. (100-150, 150-200...)
Assumed Mean $a = 225$. $\Sigma f_i = 25$. $\Sigma f_i d_i = -350$.
Mean = $225 + (-350/25) = 225 - 14 =$ ₹ 211.
Q7. Concentration of SO₂ in air. (0.00-0.04...)
Direct Method: $\Sigma f_i = 30$. $\Sigma f_i x_i = 2.96$.
Mean = $2.96 / 30 =$ 0.099 ppm.
Q8. Number of days a student was absent. (0-6, 6-10, 10-14...)
Note: Class sizes are unequal. Direct Method is best.
$\Sigma f_i = 40$. $\Sigma f_i x_i = 499$.
Mean = $499 / 40 =$ 12.48 days.
Q9. Literacy rate (in %) of 35 cities.
Assumed Mean $a = 70$. $\Sigma f_i d_i = -20$. $\Sigma f_i = 35$.
Mean = $70 + (-20/35) = 70 - 0.57 =$ 69.43 %.
Exercise 13.2 (Finding the Mode)
💡 Mode Formula: Mode = $l + \left[ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h$
Q1. Ages of patients admitted in a hospital. Find Mode and Mean.
Solution:
Max frequency is 23 in class 35-45. Modal class = 35-45.
$l = 35, f_1 = 23, f_0 = 21, f_2 = 14, h = 10$.
Mode = $35 + [ (23 - 21) / (2(23) - 21 - 14) ] \times 10$
Mode = $35 + [ 2 / (46 - 35) ] \times 10 = 35 + (20 / 11) = 35 + 1.81 =$ 36.8 years.
Mean (by direct method) = $2830 / 80 =$ 35.37 years.
Q2. Lifetimes of 225 electrical components. Find modal lifetimes.
Max frequency = 61. Modal class = 60-80.
$l = 60, f_1 = 61, f_0 = 52, f_2 = 38, h = 20$.
Mode = $60 + [ (61 - 52) / (122 - 52 - 38) ] \times 20$
Mode = $60 + [ 9 / 32 ] \times 20 = 60 + 5.625 =$ 65.625 hours.
Q3. Monthly expenditure of 200 families. Find Mode and Mean.
Modal class = 1500-2000. $f_1 = 40, f_0 = 24, f_2 = 33, h = 500$.
Mode = $1500 + [16 / 23] \times 500 =$ ₹ 1847.83.
Mean (Assumed Mean) = ₹ 2662.50.
Q4. State-wise teacher-student ratio. Find Mode and Mean.
Modal class = 30-35. $f_1 = 10, f_0 = 9, f_2 = 3, h = 5$.
Mode = $30 + [1 / 8] \times 5 =$ 30.625.
Mean = 29.2.
Q5. Runs scored by top batsmen. Find Mode.
Modal class = 4000-5000. $f_1 = 18, f_0 = 4, f_2 = 9, h = 1000$.
Mode = $4000 + [14 / 23] \times 1000 =$ 4608.7 runs.
Q6. Number of cars passing. Find Mode.
Modal class = 40-50. $f_1 = 20, f_0 = 12, f_2 = 11, h = 10$.
Mode = $40 + [8 / 17] \times 10 = 40 + 4.7 =$ 44.7 cars.