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(C) To find the standard deviation,we first calculate the midpoints $(y_i)$ and use the assumed mean method $(A = 25)$:            Class      $f_i$

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$9$

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(C) To find the standard deviation,we first calculate the midpoints $(y_i)$ and use the assumed mean method $(A = 25)$:            Class      $f_i$      $y_i$      $d_i = y_i - 25$      $f_i d_i$      $f_i d_i^2$              $0-10$      $1$      $5$      $-20$      $-20$      $400$              $10-20$      $3$      $15$      $-10$      $-30$      $300$              $20-30$      $4$      $25$      $0$      $0$      $0$              $30-40$      $2$      $35$      $10$      $20$      $200$              Total      $N = 10$      -      -      $\sum f_i d_i = -30$      $\sum f_i d_i^2 = 900$      The variance $(\sigma^2)$ is given by:$\sigma^2 = \frac{\sum f_i d_i^2}{N} - \left( \frac{\sum f_i d_i}{N} \right)^2$$\sigma^2 = \frac{900}{10} - \left( \frac{-30}{10} \right)^2$$\sigma^2 = 90 - (-3)^2 = 90 - 9 = 81$Therefore,the standard deviation $\sigma = \sqrt{81} = 9$.

$X$	$35$	$54$	$52$	$53$	$56$	$58$	$52$	$50$	$51$	$49$
$Y$	$108$	$107$	$105$	$105$	$106$	$107$	$104$	$103$	$104$	$101$

What is the standard deviation of the following series?

Class $0-10$ $10-20$ $20-30$ $30-40$

Frequency $1$ $3$ $4$ $2$

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Class Interval $0-10$ $10-20$ $20-30$ $30-40$
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Let $x_1, x_2, \dots, x_n$ be $n$ observations such that $\sum x_i^2 = 300$ and $\sum x_i = 60$. Which of the following is a possible value for $n$?

From the prices of shares $X$ and $Y$ below,find out which is more stable in value:

$X$ $35$ $54$ $52$ $53$ $56$ $58$ $52$ $50$ $51$ $49$

$Y$ $108$ $107$ $105$ $105$ $106$ $107$ $104$ $103$ $104$ $101$

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$x_i$ $4$ $3$ $1$
$f_i$ $1$ $3$ $5$

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Class Interval	$0-10$	$10-20$	$20-30$	$30-40$
Frequency	$2$	$3$	$4$	$1$

What is the standard deviation of the following series? Class $0-10$ $10-20$ $20-30$ $30-40$ Frequency $1$ $3$ $4$ $2$