Mean, variance and standard deviation using short-cut method Height in cms 70-75 75-80 80-85 85-90 9

Hindi

13.Statistics

hard

Find the mean, variance and standard deviation using short-cut method

Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$

No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$

$10.27$

Solution

Class Interval	Frequency ${f_i}$	Mid-point ${f_i}$	${y_i} = \frac{{{x_i} – 92.5}}{5}$	${y_i}^2$	${f_i}{y_i}$	${f_i}{y_i}^2$
$70-7$	$3$	$72.5$	$-4$	$16$	$-12$	$48$
$75-80$	$4$	$77.5$	$-3$	$9$	$-12$	$36$
$80-85$	$7$	$82.5$	$-2$	$4$	$-14$	$28$
$85-90$	$7$	$87.5$	$-1$	$1$	$-7$	$7$
$90-95$	$15$	$92.5$	$0$	$0$	$0$	$0$
$95-100$	$9$	$97.5$	$1$	$1$	$9$	$9$
$100-105$	$6$	$102.5$	$2$	$4$	$12$	$24$
$105-110$	$6$	$107.5$	$3$	$9$	$18$	$54$
$110-115$	$3$	$112.5$	$4$	$16$	$12$	$48$
	$60$				$6$	$254$

Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h$

$ = 92.5 + \frac{6}{{60}} \times 5 = 92.5 + 0.5 = 93$

Variance, $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 – {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} } \right]$

$=\frac{(5)^{2}}{(60)^{2}}\left[60 \times 254-(6)^{2}\right]$

$=\frac{25}{3600}(15204)=105.58$

$\therefore$ Standard deviation $(\sigma)=\sqrt{105.58}=10.27$

Standard 11

Mathematics

If $M.D.$ is $12$, the value of $S.D.$ will be

easy

The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three observations are $1, 2$ and $6$, the other two observations are

medium

The frequency distribution:

$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$

where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.

medium

If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is

$X$ $c$ $2c$ $3c$ $4c$ $5c$ $6c$

$f$ $2$ $1$ $1$ $1$ $1$ $1$

medium

(JEE MAIN-2024)

The sum of $100$ observations and the sum of their squares are $400$ and $2475$, respectively. Later on, three observations, $3, 4$ and $5$, were found to be incorrect . If the incorrect observations are omitted, then the variance of the remaining observations is

hard

(JEE MAIN-2017)

Height in cms	$70-75$	$75-80$	$80-85$	$85-90$	$90-95$	$95-100$	$100-105$	$105-110$	$110-115$
No. of children	$3$	$4$	$7$	$7$	$15$	$9$	$6$	$6$	$3$

Find the mean, variance and standard deviation using short-cut method Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$ No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$

Solution

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Find the mean, variance and standard deviation using short-cut method

Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$

No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$

The frequency distribution:

$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$

where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.

If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is

$X$ $c$ $2c$ $3c$ $4c$ $5c$ $6c$

$f$ $2$ $1$ $1$ $1$ $1$ $1$