Direct Method
The given distribution is

Here, \(\sum { f_{ i } } =64\) and \(\sum { f_{ i }x_{ i } } =848\)
Mean \(=\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\frac { 848 }{ 64 } =13.25\)

The cumulative frequencies with their respective class intervals are as follows

Cumulative frequency just greater than n/2 (i.e 30/2 = 15) is 19, belonging to class interval 55 − 60.
Median class = 55 − 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
\(\text { Median }=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h\)
= 55+((15-13)/6)x x5
= 55+10/6
= 56.67
Therefore median weight is 56.67 kg

The cumulative frequency for the given data is calculated as follows

From the table, it can be observed that n = 60
45 + x + y = 60
x + y = 15 (1)
Median of the data is given as 28.5 which lies in interval 20 - 30.
Therefore, median class = 20 - 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
\(\text { Median }=l+\left(\frac{\left(\frac{n}{2}\right)-c f}{f}\right) \times h\)
28.5 = 20 + [(60/2-(5+x))/20]xx10
8.5 = ((25-x)/2)
17 = 25 - x
8 + y =  15
y = 7
Hence, the values of x and y are 8 and 7 respectively.

From the given data, it can be observed that the maximum class frequency is 20, belonging to 40 − 50 class intervals.
Therefore, modal class = 40 − 50
Lower limit (l) of modal class = 40
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 12
Frequency (f₂) of class succeeding modal class = 11
Class size = 10
\(\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \)
\(=40+\left[\frac{20-12}{2(20)-12-11}\right] \times 10\)
= 40+((80)/(40-23))

= 40 + 4.7
= 44.7
Therefore, mode of this data is 44.7 cars.

To find the class marks, the following relation is used.
\(x_{i}=\frac{\text { Upper class limit + Lower class limit }}{2}\)
Class size (h) for this data = 10
Taking 70 as assumed mean (a), d_i, u_i, and f_iu_i are calculated as follows.

From the table, we obtain
\(\sum f_{i}=35 \)
\(\sum f_{i} u_{i}=-2 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h\)
\(=70+\left(-\frac{2}{35}\right) \times(10) \)
\(=70-\frac{20}{35} \)
\(=70-\frac{4}{7}\)
= 70 - 0.57
= 69.43
Therefore, mean literacy rate is 69.43%.

To find the class mark of each interval, the following relation is used.
\(x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}\)
Taking 17 as assumed mean (a), d_i and f_id_i are calculated as follows.

From the table, we obtain

\(\sum f_{i}=40 \)
\(\sum f_{i} u_{i}=-181 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h\)
\(=17+\left(\frac{-181}{40}\right)\)
= 17 - 4.525
= 12.475
=12.48
Therefore, the mean number of days is 12.48 days for which a student was absent.

The abscissa of point of intersection gives the median of the data.
12.5

To find median marks, we convert the given data into continuous grouped frequency distribution

\(\text { Here, } \quad \frac{n}{2}=\frac{50}{2}=25 \)
\(\therefore \text { Median class }=20-30 \)
\(\text { Median } =l+\frac{\frac{n}{2}-c f}{f} \times h=20+\frac{25-10}{20} \times 10 \)
\(=20+7.5=27.5 \)

28

35