Benefits of regular medical check-up are
(t) It enables us to take pre-medical action on time.
(it) It reduces the chances of getting sick.

Use the formula, median = \(l+\left\{\frac{\frac{n}{2}-c f}{f}\right\} \times h\)
Ans. 68

Firstly, we make the frequency distribution of the given data and then proceed to calculate mean by computing class marks (x_i), u's and f_iu_i 's as follows

Here, assumed mean, a = 45 (2) and class width, h = 10..By step deviation method,
Mean \(\left( \overline { x } \right) =a+\left\{ \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right\} \times h=45+\left\{ \frac { 630 }{ 1000 } \right\} \times 10\)
=45+63=51.3
Hence, the required mean age is 51.3 yr.

The cumulative frequencies with their respective class intervals are as follows.

It can be observed that the cumulative frequency just greater than  n/2 (i.e 400/2 = 200) is 216
belonging to class interval 3000 − 3500.
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
\(\text { Median }=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \)
\(=3000+\left(\frac{200-130}{86}\right) \times 500 \)
\(=3000+\frac{70 \times 500}{86}\)
= 3406.976
Therefore, median life time of lamps is 3406.98 hours.

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.

It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 − 35.
Therefore, modal class = 30 − 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding modal class = 9
Frequency (f₂) of class succeeding modal class = 3
\(\text { Mode }=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \times h\right) \)
\(=30+\left(\frac{10-9}{2}(10)-9-3\right) \times(5) \)
\(=30+\left(\frac{1}{20-12}\right) 5\)
= 30 + 5/8 = 30.625
Mode = 30.6
It represents that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Taking 32.5 as assumed mean (a), d_i, u_i, and f_iu_i are calculated as follows.

\(\text { Mean, } \bar{x}=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) \times h \)
\(=32.5+\left(\frac{-23}{35}\right) \times 5\)

= 32.5 - 23/7 = 32.5 - 3.28
= 29.22
Therefore, mean of the data is 29.2.
It represents that on an average, teacher−student ratio was 29.2

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%.

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore,1/2  has to be added to the upper class limit and1/2 has to be subtracted from the lower class limit of each interval.
Class mark (x_i) can be obtained by using the following relation.
\(x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}\)
Class size (h) of this data = 3
Taking 57 as assumed mean (a), d_i, u_i, f_iu_i are calculated as follows.

It can be observed that
\(\sum f_{i}=400 \)
\(\sum f_{i} u_{i}=25 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{1} u_{i}}{\sum f_{i}}\right) x h \)
\(=57+\left(\frac{25}{400}\right) \times 3\)
= 57+3/16 =  57+ 0.1875
= 57.1875
= 57.19
Mean number of mangoes kept in a packing box is 57.19.
Step deviation method is used here as the values of f_i, d_i are big and also, there is a common multiple between all d_i.

RS.145.20

Convert the given distribution into continuous grouped frequency distribution and find the median.
60.85 kg

 Here, \(\frac { n }{ 2 } =\frac { 25 }{ 2 } =12.5\)
Since, cumulative frequency just greater than 12.5 and corresponding class is 60-70.
Median class=60-70
Since, highest frequency is 9 and corresponding class is 50-60
Modal class=50-60

\(Since, sum of frequencies =181 \therefore \quad x+15+18+30+50+48+x=181 \)
\(\Rightarrow \quad 2 x=181-160=21 \Rightarrow x=10\)
Now, maximum frequency is 50. So, modal class is 13-15
\(\therefore \quad l=13, f_{1}=50, f_{0}=30, f_{2}=48 \text { and } h=2\)
\(\text { Now, Mode }=13+\frac{50-30}{100-30-48} \times 2=13+\frac{20}{22} \times 2 \)
\(=13+182=1482\)

28

Here, class interval are not continuous. We will solve it without making continuous.

Mean \(\left( \overline { x } \right) =a+\frac { \sum { f_{ i }d_{ i } } }{ \sum { f_{ i } } } =70+\frac { 9 }{ 25 } =70.36\)

Mean \(\left( \overline { x } \right) =\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\frac { 1290 }{ 50 } =25.8\)