Statistics

(d) 100

Percent frequency is $\dfrac{f}{N}\times 100$ for each class. Adding over all classes gives $\dfrac{\sum f}{N}\times 100 = 100$.

(b) The numbers 1, 5, 6, 10 have the largest possible standard deviation.

Standard deviation is largest when the values are as far from the mean as possible, which is the set 1, 2, 9, 10, so (a) is true and (b) is false (1, 5, 6, 10 are more bunched towards the centre). It is smallest when the values are as close together as possible, i.e. four consecutive numbers such as 4, 5, 6, 7 or 7, 8, 9, 10, which have the same spread, so (c) and (d) are true.

Arrange the values of $x$ in ascending order with their frequencies and form the cumulative frequency.

$N = 59$, so $N + 1 = 60$.

$Q_1$: position $= \dfrac{N+1}{4} = 15$th item. The first $cf$ not less than 15 is 23, so $Q_1 = 6$.

$D_6$: position $= \dfrac{6(N+1)}{10} = 36$th item. The first $cf$ not less than 36 is 45, so $D_6 = 10$.

$P_{76}$: position $= \dfrac{76(N+1)}{100} = 45.6$th item, taken as the 46th. The first $cf$ not less than 45.6 is 53, so $P_{76} = 12$.

The more consistent student is the one with the smaller coefficient of variation.

Ram ($N=10$): $\sum x = 610$, so $\bar x = 61$.

Deviations $(x-\bar x)$: $-7, -1, -5, 7, 11, -9, -13, 15, 19, -17$, giving $\sum (x-\bar x)^2 = 1370$. $\sigma = \sqrt{\frac{1370}{10}} = \sqrt{137} \approx 11.71$ $CV = \frac{11.71}{61}\times100 \approx 19.19\%$

Hari ($N=10$): $\sum y = 615$, so $\bar y = 61.5$.

Deviations: $4.5, -4.5, -10.5, 10.5, 7.5, 1.5, -1.5, -7.5, 13.5, -13.5$, giving $\sum (y-\bar y)^2 = 742.5$. $\sigma = \sqrt{\frac{742.5}{10}} = \sqrt{74.25} \approx 8.62$ $CV = \frac{8.62}{61.5}\times100 \approx 14.01\%$

Since $CV_{\text{Hari}} (14.01\%) < CV_{\text{Ram}} (19.19\%)$, Hari has the more consistent performance.

(d) Any of the above averages

Mean deviation is the average of the absolute deviations of the observations from an average, and that average may be the mean, the median or the mode — so it can be taken from any of these averages.

Take assumed mean $A=25$ and class width $h=10$, with $d=\dfrac{x-25}{10}$.

$\bar x=A+h\frac{\sum fd}{N}=25+10\left(\frac{14}{50}\right)=27.8$

$\sigma=h\sqrt{\frac{\sum fd^2}{N}-\left(\frac{\sum fd}{N}\right)^2}=10\sqrt{1.6-0.0784}=10\sqrt{1.5216}\approx12.34$

$CV=\frac{\sigma}{\bar x}\times100=\frac{12.335}{27.8}\times100\approx44.37\%$

Standard deviation $\approx12.34$ marks and coefficient of variation $\approx44.37\%$.

Given: $N = 50$, so $\dfrac{N}{4} = 12.5$ and $\dfrac{3N}{4} = 37.5$

First quartile ($Q_1$): $cf \ge 12.5$ first occurs in the class 30–40, so the $Q_1$ class is 30–40.

$L = 30,\ cf_p = 11,\ f = 12,\ h = 10,\ \frac{N}{4} = 12.5$

$Q_1 = L + \frac{\frac{N}{4} - cf_p}{f} \times h = 30 + \frac{12.5 - 11}{12} \times 10 = 30 + \frac{15}{12} = 31.25$

Third quartile ($Q_3$): $cf \ge 37.5$ first occurs in the class 40–50, so the $Q_3$ class is 40–50.

$L = 40,\ cf_p = 23,\ f = 15,\ h = 10,\ \frac{3N}{4} = 37.5$

$Q_3 = L + \frac{\frac{3N}{4} - cf_p}{f} \times h = 40 + \frac{37.5 - 23}{15} \times 10 = 40 + \frac{145}{15} = 49.67$

Quartile deviation:

$QD = \frac{Q_3 - Q_1}{2} = \frac{49.67 - 31.25}{2} = \frac{18.42}{2} = 9.21$

Coefficient of quartile deviation:

$\text{Coefficient of } QD = \frac{Q_3 - Q_1}{Q_3 + Q_1} = \frac{49.67 - 31.25}{49.67 + 31.25} = \frac{18.42}{80.92} = 0.228$

Result: Quartile deviation = 9.21 and its coefficient = 0.228 (approximately).

Given: $N=\sum f=60$.

Mean: $\bar x=\frac{\sum fx}{N}=\frac{1640}{60}=27.33$

Variance: $\sigma^2=\frac{\sum fx^2}{N}-\left(\frac{\sum fx}{N}\right)^2=905-(27.33)^2=905-747.11=157.89$

Standard deviation: $\sigma=\sqrt{157.89}\approx12.57$

Interpretation: the marks/values are, on average, about 12.57 units away from the mean of 27.33. The coefficient of variation is $CV=\frac{12.57}{27.33}\times100\approx46\%$ so the distribution shows moderate to high variability.