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Answer 1

$\large\underline{\sf{Solution-}}$

Calculation of Mean using Step Deviation Method

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c|c|c}\sf Class\: interval&\sf Frequency\: (f_i)&\sf \: midvalue \: (x_i)&\sf \: u_i&\sf \: f_iu_i\\\frac{\qquad  \qquad}{}&\frac{\qquad  \qquad}{}\\\sf 0 - 20&\sf 17&\sf10&\sf - 1&\sf - 17\\\\\sf 20 - 40 &\sf 2&\sf30 - A&\sf 0&\sf 0\\\\\sf 40-60 &\sf 32 &\sf50&\sf1&\sf32\\\\\sf 60 - 80&\sf y&\sf70&\sf2&\sf2y\\\\\sf 80-100&\sf 19&\sf90&\sf3&\sf57\\\frac{\qquad}{}&\frac{\qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

Given that,

$\rm \: \sum \: f_i = 120$

$\rm \: 68 + x + y = 120$

$\rm \: x + y = 120 - 68$

$\rm\implies \:x + y = 52 - - - - (1)$

Now, From frequency distribution table, we have

$\rm \: A = 30$

$\rm \: \sum \: f_i u_i= 2y + 72$

$\rm \: h = 20$

$\rm \: Mean = 50$

Now, We know, Mean using Step Deviation Method is given by

$\boxed{\tt{ \: \: Mean \: = \: A \: + \: h \: \times \: \dfrac{\sum f_iu_i}{\sum f_i} \: \: }}$

So, on Substituting all the values, we get

$\rm \: 50 = 30 + 20 \times \dfrac{72 + 2y}{120}$

$\rm \: 50 - 30 = \dfrac{72 + 2y}{6}$

$\rm \: 20 = \dfrac{72 + 2y}{6}$

$\rm \: 72 + 2y = 120$

$\rm \: 2y = 120 - 72$

$\rm \: 2y = 48$

$\rm\implies \:y = 24$

On substituting the value of y in equation (1), we get

$\rm \:x + 24 = 52$

$\rm \:x = 52 - 24$

$\rm\implies \:x = 28$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\bf{x = 28} \\ \\ &\bf{y = 24} \end{cases}\end{gathered}\end{gathered}$

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ADDITIONAL INFORMATION

1. Mean using Direct Method is given by

$\boxed{\tt{ \: \: Mean \: or\: \sf \overline{x} = \dfrac{\sum f_ix_i}{\sum f_i} \: \: }}$

2. Mean Using Short Cut Method is given by

$\boxed{\tt{ \: \: Mean \: or \: \sf \overline{x} = A + \dfrac{\sum f_id_i}{\sum f_i} \: \:where \: d \: = \: x_i \: - \: A}}$

Answer 2

Given :-

$\begin{center}\begin{tabular}{ c c c } \sf Class & \sf Frequency\\\sf 0-20 &\sf 17\\ \sf 20-40&\sf x\\\sf 40-60&\sf 32\\\sf 60-80&\sf y\\\sf 80-100&\sf 19 \\\sf Total&\sf 120 \end{tabular}\end{center}$

To Find :-

Value of x and y

Solution :-

$\begin{center}\begin{tabular}{ c c c } \sf Class & \sf Frequency(x_1)&Mid-Value(y_1)\\\sf 0-20 &\sf 17&\sf 10\\ \sf 20-40&\sf x&30\\\sf 40-60&\sf 32&\sf 50\\\sf 60-80&\sf y&\sf 70\\\sf 80-100&\sf 19&\sf 90 \\\sf &\sf 68 + x + y = 120 \end{tabular}\end{center}$

Since,

68 + x + y = 120

x + y = 120 - 68

x + y = 52 (1)

Now Finding Sum

$\begin{center}\begin{tabular}{ c c c } \sf \sf Frequency(x_1)&Mid-Value(y_1)& \sum \sf f_ix_i\\\sf 17&\sf 10&\sf 170\\\sf x&30&\sf 30x\\\sf 32&\sf 50&\sf 1600\\\sf y&\sf 70&\sf 70y\\\sf 19&\sf 90&\sf 1710 \\\ &&\sf Total = 3480+30x+70y \end{tabular}\end{center}$

Now

Mean = Σfixi/Σfi

50 = 3480 + 30x + 70y/120

50 × 120 = 3480 + 30x + 70y

6000 = 3480 + 30x + 70y

6000 - 3480 = 30x + 70y

2520 = 30x  + 70y

2520/10 = 30x/10 + 70y/10

252 = 3x + 7y (2)

On multiplying Eq. 1 by 3

3(x + y) = 3(52)

3x + 3y = 156 (3)

On subtracting 2 and 3

156 - 252 = 3x + 3y - 3x - 7y

-96 = 3y - 7y

-96 = -4y

96 = 4y

96/4 = y

24 = y

From 1

x + y = 52

x + 24 = 52

x = 52 - 24

x = 28