MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass
        The mean and standard deviation of 50 items is found tobe 40 and 5 respectively. It was later found that oneitem 50 was incorrect and the correct figure is 40.Calculate the correct standard deviation. 5
9 months ago

Answers : (1)

Yasas Chandra Sai P
17 Points
							
Mean = 40
Standard deviation = 5
Number of items =50
Name the items as a1,a2,,.......a50
Suppose the 50th item is wrongly indicated as 50 instead of 40
Sum of all observations before correction = 50 x 40 = 2000
Sum of the correct observations (a1+a2....a49) = 2000-50 = 1950
Sum of 50 observations = 1950 + corrected observation = 1950 + 40 = 1990
Mean = 1990 / 50 = 199/5 =39.8
Sum of the squares of the observations = (a1^2+a2^2-------a50^2) = n(SD^2+Mean^2) = 50(4^2+40^2)
(a1^2+a2^2-----------a49^2)+(50)^2 = 50(4^2+40^2)
(a1^2+a2^2+----------a49^2) = 50(16+1600) – 2500
Corrected sum of squares = (a1^2+a2^2+----------------a49^2)+(40^2) = 50(1616)-2500+1600 = 79,900
Mean of squares = 79,900 / 50 = 1598
SD^2 =Mean of squares – Corrected Mean^2 = 1598-(39.8)^2 = 13.96
SD = (13.96)^1/2  
 
  
9 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Get Extra Rs. 233 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details