Search Colleges, Courses, Exams, Questions and Articles

Ask a query on 8879100846 Ask Login | Sign Up

 

 

› Ask & Answer › Question

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

0 3 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    2 months ago

    This is a long answer type question as classified in NCERT Exemplar

    G i v e n     t h a t     n = 1 0 0 ,     x ¯ = 4 0 ,     σ = 1 0 ∴     x ¯ = ∑ x i n                   ⇒ 4 0 = ∑ x i 1 0 0           ⇒ ∑ x i = 4 0 0 0 C o r r e c t e d     ∑ x i = 4 0 0 0 − 3 0 − 7 0 + 3 + 2 7 = 3 9 3 0 a n d     C o r r e c t e d     m e a n = 3 9 3 0 1 0 0 = 3 9 . 3 N o w         σ 2 = ∑ x i 2 n − ( 4 0 ) 2         ⇒           1 0 0 = ∑ x i 2 1 0 0 − 1 6 0 0 ⇒         ∑ x i 2 = 1 7 0 0 × 1 0 0                     ⇒ ∑ x i 2 = 1 7 0 0 0 0 ∴ C o r r e c t e d     ∑ x i 2 = 1 7 0 0 0 0 − ( 3 0 ) 2 − ( 7 0 ) 2 + ( 3 ) 2 + ( 2 7 ) 2                                                                                 = 1 7 0 0 0 0 − 9 0 0 − 4 9 0 0 + 9 + 7 2 9 = 1 6 4 9 3 8 ∴     C o r r e c t     S D = 1 6 4 9 3 8 1 0 0 − ( 3 9 . 3 ) 2 = 1 6 4 9 . 3 8 − 1 5 4 4 . 4 9 = 1 0 4 . 8 9 = 1 0 . 2 4 H e n c e ,     t h e     r e q u i r e d     S D = 1 0 . 2 4

Similar Questions for you

Given data 60, 60, 44, 58, 68, α, β, 56 has mean 58, variance = 66.2 then find α2 + β2
A
alok kumar singh

Variance = ∑ x 2 n − ( x ¯ ) 2  

6 0 2 + 6 0 2 + 4 4 2 + 5 8 2 + 6 8 2 + α 2 + β 2 + 5 6 2 8 = ( 5 8 ) 2 = 6 6 . 2            

7 2 0 0 + 1 9 3 6 + 3 3 6 4 + 4 6 2 4 + 3 1 3 6 + α 2 + β 2 8 = 3 3 6 4 = 6 6 . 2             

2 5 3 2 . 5 + α 2 + β 2 8 − 3 3 6 4 = 6 6 . 2            

α2 + β2 = 897.7 × 8

= 7181.6

For the following data table

xi

fi

0 – 4

4 – 8

8 – 12

12 – 16

16 – 20

2

4

7

8

6

Find the value of 20 M (where M is median of the data)
A
alok kumar singh

xi

fi

c.f.

0 – 4

4 – 8

8 – 12

12 – 16

16 – 20

2

4

7

8

6

2

6

13

21

27

N = ∑ f = 2 7

( N 2 ) = 2 7 2 = 1 3 . 5

So, we have median lies in the class 12 – 16

I1 = 12, f = 8, h = 4, c.f. = 13

So, here we apply formula

M = I 1 + N 2 − c . f . f × h = 1 2 + 1 3 . 5 − 1 3 8 × 4

= 1 2 + 5 2

M = 2 4 . 5 2 = 1 2 . 2 5

20 M = 20 × 12.25

= 245

Let mean and variance of 6 observations a, b, 68, 44, 40, 60 be 55 and 194. If a > b then find a + 3b
A
alok kumar singh

  a + b + 6 8 + 4 4 + 4 0 + 6 0 6 = 5 5

212 + a + b = 330

⇒ a + b = 118

∑ x i 2 n − ( x ¯ ) 2 = 1 9 4          

a 2 + b 2 + ( 6 8 ) 2 + ( 4 4 ) 2 + ( 4 0 ) 2 + ( 6 0 ) 2 6 = ( 5 5 ) 2 = 1 9 4

= 3219

11760 + a2 + b2 = 19314

⇒ a2 + b2 = 19314 – 11760

= 7554

(a + b)2 –2ab = 7554

From here b = 41.795

a + b = 118

⇒ a + b + 2b = 118 + 83.59

= 201.59

Let X1, X2,…..X18 be eighteen observations such that ∑ i = 1 1 8 ( X i − α ) = 3 6     a n d     ∑ i = 1 1 8 ( X i − β ) 2 = 9 0 ,  where a and b are distinct real numbers. If the standard deviation of these observations is 1, then the value of   | α − β |     i s . . . . . . . .

A
alok kumar singh

Kindly go throuigh the solution

Given   ∑ i = 1 1 8 ( x i − α ) = 3 6     i . e         ∑ i = 1 1 8 x i − 1 8 α = 3 6 . . . . . . . . . . ( i )

&       ∑ i = 1 1 8 ( x i − β ) 2 = 9 0         i . e ∑ i = 1 1 8 x i 2 − 2 β ∑ x i + 1 8 β 2 = 9 0 . . . . . . . . . . . . . ( i i )      

(i) & (ii)   ∑ i = 1 1 8 x i 2 = 9 0 − 1 8 β + 3 6 β ( α + 2 ) . . . . . . . . . . . . . ( i i i )

Now variance σ 2 = ∑ x i 2 n − ( ∑ x i n ) 2 = 1 given

->(a - b) (a - b + 4) = 0

Since α ≠ β     s o     | α − β | = 4  

 

The number of values of a ∈ N  such that the variance of 3, 7, 12, a, 43 -a is a natural number is:
V
Vishal Baghel

M e a n = 3 + 1 2 + 7 + a + ( 4 3 − a ) 5 = 1 3  

Variance = 3 2 + 1 2 2 + 7 2 + a 2 + ( 4 3 − a ) 2 5 − ( 1 3 ) 2  

⇒ 2 a 2 − a + 1 5 → N a t u r a l     n u m b e r      

Let 2a2 – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not 4 λ     o r     4 λ + 1     f r o m .  

As each square form is 4 λ     o r     4 λ + 1  

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Write a Review

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Ask Now
Community GuidelinesUser Points System
QuestionsDiscussionsTags
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post