Let the assumed mean, a=55 and width of class interval, h=10.Table for given data is

Now, we have, N=90+x+y
\(\Rightarrow \quad 120=90+x+y\)
\(\Rightarrow \quad x+y=30\)
\(\because \quad Mean=59\)
\(\Rightarrow \quad a+\left( \frac { \sum { f_{ i }u_{ i } } }{ N } \right) \times h=59\)
\(\Rightarrow \quad 55+\left( \frac { 44+4y }{ 120 } \right) \times 10=59\)
\(\Rightarrow \quad \frac { 11+y }{ 3 } =4\)
\(\Rightarrow \quad 11+y=12\)
\(\Rightarrow \quad y=1\)
On putting the value of y in Eq.(i), we get
x + 1 = 30⇒ x = 29
Hence, x = 29 and y = 1.

Mode = \(l+\frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(l+\frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(=45+\frac { 2 }{ 18 } x10\)
=46.1

a = Assumed mean = 12
Mean \(\overset { - }{ =a+x } \frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
Mean =12+\(\frac { -366 }{ 400 } \times 2=10.17\)
\(\frac { \Sigma f }{ 2 } =200\Rightarrow \) Median class=09-11

34, 46

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.

Table for the given data is

Here, a = 18, N = 44 + f and \(\sum\)f_id_i = 2f - 40
\(\begin{aligned} \therefore \quad \operatorname{Mean}(\bar{x}) & =a+\frac{1}{N} \times \Sigma f_i d_i \\ \end{aligned}\)
\(\begin{aligned} =18+\frac{1}{(44+f)} \times(2 f-40) \end{aligned}\) ...(i)
But mean = 18               [given] ...(ii)
On equating Eqs. (i) and (ii), we get
\(\begin{array}{rlrl} 18 & =18+\frac{1}{(44+f)} \times(2 f-40) \\ \end{array}\)
\(\begin{array}{rlrl} & \Rightarrow & 0 & =\frac{2 f-40}{44+f} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & 2 f-40 =0 \Rightarrow 2 f=40 \\ \end{array}\)
\(\begin{array}{rlrl} \therefore f =\frac{40}{2}=20 \end{array}\)
Hence, the missing frequency is 20.

From the table
12+a=25 \(\Rightarrow \) a=25-12=13
25+10=b \(\Rightarrow \) b=35
b+c=43 \(\Rightarrow \) c=43-b=43-35 =8
48+2=d \(\Rightarrow \) d=50

Modal Class = 80- 100
i=80 , f₁=18 , f₂=5,f₀=12,h=20
Model = i + \(\left( \frac { f_{ i }-f_{ 0 } }{ 2f_{ i }-f_{ 0 }-f_{ 2 } } \right) \times 20\)
\(=80+\left( \frac { 18-12 }{ 36-12-5 } \right) \times 20\)
\(=80+\frac { 6 }{ 19 } \times 20\)
=80+6.31
=86.31

Here, we have the cumulative frequency distribution of less than type. We observe that the number of students getting marks less than 10 is 14 and 22 students have marks less than 20.
Therefore, number of students getting marks between 10 and 20 is 22-14=8. Similarly, the number of students getting marks between 20 and 30 is 37-22=15 and so on.
Thus, we have the following frequency distribution table