Frequency \left(f\right)< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 4< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 24< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 28< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > \alpha< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 8< / t d > < / t r > < / t b o d y > < / t a b l e > < p > b e </td> </tr> </tbody></table><p>be < / t d >< / t r >< / t b o d y >< / t ab l e >< p > b e 5 .I f m a n d If m and I f man d {\sigma}^{2} are respectively the mean deviation about the mean and the variance of the data, then 3 α m + σ 2 \dfrac{3\alpha}{m + {\sigma}^{2}} m + σ 2 3 α is equal to _______ .</p>
Step-by-step solution
Given,
The mean of the below data is 5 5 5
x x x 1 1 1 3 3 3 5 5 5 7 7 7 9 9 9 Frequency \left(f\right)< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 4< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 24< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 28< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > \alpha< / t d > < t d s t y l e = " t e x t − a l i g n : c e n t e r ; " > </td> <td style="text-align: center;"> < / t d >< t d s t y l e = " t e x t − a l i g n : ce n t er ; " > 8< / t d > < / t r > < / t b o d y > < / t a b l e > < p > N o w f i n d i n g m e a n w e g e t , < / p > < p > </td> </tr> </tbody></table><p>Now finding mean we get,</p><p> < / t d >< / t r >< / t b o d y >< / t ab l e >< p > N o w f in d in g m e an w e g e t , < / p >< p > \dfrac{4 + 72 + 28 \times 5 + 7\alpha + 72}{64 + \alpha} = 5< / p > < p > </p><p> < / p >< p > ⇒ \alpha = 16< / p > < p > S o , t o t a l n u m b e r o f o b s e r v a t i o n w i l l b e , </p><p>So, total number of observation will be, < / p >< p > S o , t o t a l n u mb ero f o b ser v a t i o n w i ll b e , ∑ {f}_{i} = 80< / p > < p > N o w f i n d i n g , m e a n d e v i a t i o n w e g e t , < / p > < p > </p><p>Now finding, mean deviation we get,</p><p> < / p >< p > N o w f in d in g , m e an d e v ia t i o n w e g e t , < / p >< p > M . D = \dfrac{∑ {f}_{i}\left|{x}_{i} - 5\right|}{∑ {f}_{i}}< / p > < p > </p><p> < / p >< p > = \dfrac{4 \times 4 + 24 \times 2 + 0 + 16 \times 2 + 8 \times 4}{80}= \dfrac{8}{5}< / p > < p > S o , </p><p>So, < / p >< p > S o , m = \dfrac{8}{5}< / p > < p > N o w f i n d i n g , v a r i a n c e w e g e t , < / p > < p > </p><p>Now finding, variance we get, </p><p> < / p >< p > N o w f in d in g , v a r ian ce w e g e t , < / p >< p > {\sigma}^{2} = \dfrac{∑ {f}_{i}{\left|{x}_{i} - 5\right|}^{2}}{∑ {f}_{i}}< / p > < p > </p><p> < / p >< p > ⇒ {\sigma}^{2} = \dfrac{4 \times 16 + 24 \times 4 + 0 + 16 \times 4 + 8 \times 16}{80}< / p > < p > </p><p> < / p >< p > ⇒ {\sigma}^{2} = \dfrac{22}{5}< / p > < p > H e n c e , t h e v a l u e o f </p><p>Hence, the value of < / p >< p > He n ce , t h e v a l u eo f \dfrac{3\alpha}{m + {\sigma}^{2}} = \dfrac{3 \times 16}{\dfrac{8}{5} + \dfrac{22}{5}} = 8
Practice Statistics on Quantrex Academy →
← Previous question The mean and standard deviation of the marks of 10 students were found to be 50 and 12…Next question → Let the positive numbers a _ 1 , a _ 2, a _ 3 , a _ 4 and a _ 5 be in a G.P. Let their…
More from Statistics
Consider a data consisting of 10 observations x_1, x_2, …, x_ 10 , whose mean is 5 and… JEE Advanced 2026 A set of four observations has mean 1 and variance 13 . Another set of six observations… JEE Main 2026 Let the mean and the variance of seven observations 2, 4, α, 8, β, 12, 14 , α < β , be 8… JEE Main 2026 A data consists of 20 observations x_1, x_2, …, x_ 20 . If ∑_ i=1 ^ 20 (x_i + 5)^2 = 2500… JEE Main 2026 A variable X takes values 0, 0, 2, 6, 12, 20, …, n(n-1) with frequencies ^nC_0, ^nC_1… JEE Main 2026 The mean deviation about the mean for the data x_i 5 7 9 10 12 15 f_i 8 6 2 2 2 6 is… JEE Main 2026 For 10 observations x_1, x_2, …, x_ 10 , if ∑_ i=1 ^ 10 (x_i+2)^2=180 and ∑_ i=1 ^ 10… JEE Main 2026 Suppose that the mean and median of the non-negative numbers 21, 8, 17, a, 51, 103, b… JEE Main 2026 The mean and variance of n observations are 8 and 16 , respectively. If the sum of the… JEE Main 2026 If the mean of the data Class 5-10 10-15 15-20 20-25 25-30 30-35 Frequency 2 k 28 54 k+1… JEE Main 2026 The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these… JEE Main 2026 Let X=\ x ∈ ~N: 1 ≤ x ≤ 19\ and for some a, b ∈ ℝ, Y=\ a x+b: x ∈ X\ . If the mean and… JEE Main 2026 The mean and variance of a data of 10 observations are 10 and 2, respectively. If an… JEE Main 2026 If the mean and the variance of the data Class 4-8 8-12 12-16 16-20 Frequency 3 λ 4 7 are… JEE Main 2026 Let the mean and variance of 8 numbers -10,-7,-1, x, y, 9,2,16 be (7)/(2) and (293)/(4)… JEE Main 2026 If the mean deviation about the median of the numbers k, 2 k, 3 k, …. ., 1000 k is 500… JEE Main 2026
Full Statistics list
Statistics JEE Maths PYQs JEE Mathematics PYQs
All JEE Main PYQs
© Quantrex Academy · Free JEE Main, JEE Advanced, NEET, NDA & BITSAT previous year questions with solutions