बारम्बारता तालिका, रेखाचित्र, सहसम्बन्ध

Class width $h = 10$, total students $N = \sum f = 10+12+6+14+30+15+8+5 = 100$.

(a) Mean

$\bar{x} = \frac{\sum fm}{N} = \frac{3940}{100} = 39.4$

(b) Median

$N/2 = 50$. The first c.f. that is at least 50 is 72, so the median class is 40-50 ($L=40$, c.f. before $=42$, $f=30$).

$Md = L + \frac{N/2 - c.f.}{f}\times h = 40 + \frac{50-42}{30}\times 10 = 40 + 2.67 = 42.67$

(c) First quartile

$N/4 = 25$. The first c.f. at least 25 is 28, so the $Q_1$ class is 20-30 ($L=20$, c.f. before $=22$, $f=6$).

$Q_1 = 20 + \frac{25-22}{6}\times 10 = 20 + 5 = 25$

(d) Standard deviation

$\sigma = \sqrt{\frac{\sum fm^2}{N} - \left(\frac{\sum fm}{N}\right)^2} = \sqrt{\frac{191900}{100} - (39.4)^2} = \sqrt{1919 - 1552.36} = \sqrt{366.64} \approx 19.15$

Result: Mean = 39.4, Median = 42.67, $Q_1$ = 25, S.D. = 19.15 marks.

Let $X$ = marks in population and $Y$ = marks in health, $n = 8$.

Karl Pearson's coefficient of correlation:

$r = \frac{n\sum XY - \sum X\sum Y}{\sqrt{n\sum X^2 - (\sum X)^2}\,\sqrt{n\sum Y^2 - (\sum Y)^2}}$

Numerator:

$8(37563) - (548)(548) = 300504 - 300304 = 200$

Denominator:

$\sqrt{8(37580) - 548^2} = \sqrt{300640 - 300304} = \sqrt{336}$

and the same for $Y$, so

$r = \frac{200}{\sqrt{336}\times\sqrt{336}} = \frac{200}{336} = 0.595$

Result: $r \approx +0.60$. There is a moderate positive correlation between marks in population and marks in health: students who score well in one tend to score fairly well in the other.

Class width $h = 20$, $N = 15+20+25+15+10+5 = 90$.

(a) Average (mean) wage

$\bar{x} = \frac{\sum fm}{N} = \frac{9900}{90} = \text{Rs } 110$

(b) Median wage

$N/2 = 45$, so the median class is 100-120 ($L=100$, c.f. before $=35$, $f=25$).

$Md = 100 + \frac{45-35}{25}\times 20 = 100 + 8 = \text{Rs } 108$

(c) Mode

The highest frequency is 25, so the modal class is 100-120 ($L=100$, $f_1=25$, $f_0=20$, $f_2=15$).

$Mo = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\times h = 100 + \frac{25-20}{50-20-15}\times 20 = 100 + \frac{5}{15}\times 20 = \text{Rs } 106.67$

(d) Standard deviation

$\sigma = \sqrt{\frac{\sum fm^2}{N} - \left(\frac{\sum fm}{N}\right)^2} = \sqrt{\frac{1161000}{90} - 110^2} = \sqrt{12900 - 12100} = \sqrt{800} \approx \text{Rs } 28.28$

Result: Mean = Rs 110, Median = Rs 108, Mode = Rs 106.67, S.D. = Rs 28.28.

(Karl Pearson's coefficient of correlation)

$r=\frac{\sum xy}{N\,\sigma_x\,\sigma_y}$

where $x=X-\bar{X}$, $y=Y-\bar{Y}$, $N$ = number of pairs, and $\sigma_x,\ \sigma_y$ are the standard deviations of $X$ and $Y$.

(Equivalent direct form: $r=\dfrac{N\sum XY-(\sum X)(\sum Y)}{\sqrt{N\sum X^2-(\sum X)^2}\ \sqrt{N\sum Y^2-(\sum Y)^2}}$)

Simple line graph and multiple line graph.

(Other acceptable pairs: simple line graph and cumulative line graph; multiple line graph and component line graph.)

Calculation of Median

Median position:

$\frac{N}{2} = \frac{80}{2} = 40$

The cumulative frequency just greater than 40 is 56, so the median class is 40–50.

For the median class:

• Lower limit, $l = 40$
• Preceding cumulative frequency, $cf = 36$
• Frequency of median class, $f = 20$
• Class width, $h = 10$

$\text{Median} = l + \frac{\frac{N}{2} - cf}{f} \times h$

$= 40 + \frac{40 - 36}{20} \times 10$

$= 40 + \frac{4}{20} \times 10$

$= 40 + 2 = 42$

Median marks = 42