Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  CBSE XI  >>  Math  >>  Statistics
Answer
Comment
Share
Q)

Find the mean deviation about the mean for data : $x_i :15,21,27,30,35 \qquad f_i : 3,5,6,7,8$

$\begin{array}{1 1}(A)\;5.1\\(B)\;3\\(C)\;2.5\\(D)\;7.6\end{array} $

1 Answer

Comment
A)
Step 1:
$x_i =15,\qquad f_i=3, \qquad CF=3,\qquad |x_i -M|=|15-30|=15 \qquad f_i |x_i-M|=45$
$x_i =21,\qquad f_i=5, \qquad CF=8,\qquad |x_i -M|=|21-30|=3 \qquad f_i |x_i-M|=18$
$x_i =27,\qquad f_i=6, \qquad CF=14,\qquad |x_i -M|=|27-30|=3 \qquad f_i |x_i-M|=18$
$x_i =30,\qquad f_i=7, \qquad CF=21,\qquad |x_i -M|=|30-30|=0 \qquad f_i |x_i-M|=0$
$x_i =35,\qquad f_i=8, \qquad CF=29,\qquad |x_i -M|=|35-30|=5 \qquad f_i |x_i-M|=40$
Total $N= \sum f_i=29(odd)$
$Medium M= \bigg( \large\frac{N+1}{2} \bigg)$ th observation
$\qquad= \bigg( \large\frac{29+1}{2} \bigg)$ th value =15 th
Step 2:
Mean deviation about median $= \large\frac{\sum f_i |x_i-M|}{\sum f_i}$
$\qquad= \large\frac{148}{29}$
$\qquad= 5.1$
Hence A is the correct answer.
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...