Find the mean,variance,and standard deviation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The calculations are performed using the step-deviation method where $h = 5$ and assumed mean $A = 92.5$.      Class Interval    Frequency $f_i$

Vedclass · Accepted Answer

Mean $= 93$,Variance $= 105.58$,Standard Deviation $= 10.27$

Vedclass · Answer

(A) The calculations are performed using the step-deviation method where $h = 5$ and assumed mean $A = 92.5$.      Class Interval    Frequency $f_i$    Mid-point $x_i$    $y_i = \frac{x_i - 92.5}{5}$    $f_i y_i$    $f_i y_i^2$        $70-75$    $3$    $72.5$    $-4$    $-12$    $48$        $75-80$    $4$    $77.5$    $-3$    $-12$    $36$        $80-85$    $7$    $82.5$    $-2$    $-14$    $28$        $85-90$    $7$    $87.5$    $-1$    $-7$    $7$        $90-95$    $15$    $92.5$    $0$    $0$    $0$        $95-100$    $9$    $97.5$    $1$    $9$    $9$        $100-105$    $6$    $102.5$    $2$    $12$    $24$        $105-110$    $6$    $107.5$    $3$    $18$    $54$        $110-115$    $3$    $112.5$    $4$    $12$    $48$        Total    $N = 60$    -    -    $\sum f_i y_i = 6$    $\sum f_i y_i^2 = 254$  Mean $\bar{x} = A + \left( \frac{\sum f_i y_i}{N} ight) 	imes h = 92.5 + \left( \frac{6}{60} ight) 	imes 5 = 92.5 + 0.5 = 93$.Variance $\sigma^2 = \frac{h^2}{N^2} [N \sum f_i y_i^2 - (\sum f_i y_i)^2] = \frac{25}{3600} [60 	imes 254 - (6)^2] = \frac{25}{3600} [15240 - 36] = \frac{25}{3600} 	imes 15204 = 105.58$.Standard Deviation $\sigma = \sqrt{105.58} \approx 10.27$.

Find the mean,variance,and standard deviation using the short-cut method for the following data:

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

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

Similar Questions

The standard deviation of the following distribution is:
Class interval $0-10$ $10-20$ $20-30$ $30-40$
Frequency $1$ $3$ $4$ $2$

If the mean and standard deviation of $n$ observations $x_1, x_2, \dots, x_n$ are $\bar{x}$ and $\sigma$ respectively,then what is the sum of the squares of the observations?

The mean and standard deviation of $100$ items are $50$ and $4$ respectively. The sum of the squares of all the items is:

$A$ data consists of $20$ observations $x_1, x_2, ..., x_{20}$. If $\sum_{i=1}^{20} (x_i + 5)^2 = 2500$ and $\sum_{i=1}^{20} (x_i - 5)^2 = 100$, then the ratio of mean to standard deviation of this data is:

Given that the total of $16$ values is $528$ and the sum of the squares of deviations from $33$ is $9158$. The variance is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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

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

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