Home
Class 12
MATHS
Consider the observation x(1)=1, x(2), x...

Consider the observation `x_(1)=1, x_(2), x_(3)=3, .............x_(100)=100`
`x_(101), x_(102), x_(103), x_(104)=104` Median of the given data is

A

51

B

52.5

C

51.5

D

53

Text Solution

AI Generated Solution

The correct Answer is:
To find the median of the given data set, we will follow these steps: ### Step 1: Identify the Data Set The observations are given as: - \( x_1 = 1 \) - \( x_2 = 2 \) - \( x_3 = 3 \) - ... - \( x_{100} = 100 \) - \( x_{101} = 104 \) - \( x_{102} = 104 \) - \( x_{103} = 104 \) - \( x_{104} = 104 \) This means the data set consists of the numbers 1 through 100, followed by the number 104 repeated four times. ### Step 2: Count the Total Number of Observations The total number of observations is: - From \( x_1 \) to \( x_{100} \): 100 observations - From \( x_{101} \) to \( x_{104} \): 4 observations Thus, the total number of observations is: \[ 100 + 4 = 104 \] ### Step 3: Determine the Position of the Median Since there are 104 observations (an even number), the median will be the average of the values at the positions \( \frac{n}{2} \) and \( \frac{n}{2} + 1 \), where \( n \) is the total number of observations. Calculating these positions: - \( \frac{104}{2} = 52 \) - \( \frac{104}{2} + 1 = 53 \) ### Step 4: Find the Values at the Median Positions Now we need to find the values at the 52nd and 53rd positions in the ordered data set. The ordered data set is: - \( 1, 2, 3, \ldots, 100, 104, 104, 104, 104 \) The 52nd observation is 52, and the 53rd observation is 53. ### Step 5: Calculate the Median The median is the average of the 52nd and 53rd observations: \[ \text{Median} = \frac{x_{52} + x_{53}}{2} = \frac{52 + 53}{2} = \frac{105}{2} = 52.5 \] ### Final Answer The median of the given data set is \( 52.5 \). ---
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • STATISTICS

    FIITJEE|Exercise Assertion -Reasoning Type MCQ Single Correct(Level II)|2 Videos
  • SET, RELATION & FUNCTION

    FIITJEE|Exercise Exercise 3|8 Videos
  • STRAIGHT LINE

    FIITJEE|Exercise NUMERICAL BASED|4 Videos

Similar Questions

Explore conceptually related problems

Consider any set of 201 observations x_(1),x_(2),...,x_(200),x_(201). It is given that x_(1)

The standard deviation of n observations x_(1),x_(2),......,x_(n) is 2. If sum_(i=1)^(n)x_(i)=20 and sum_(i=1)^(n)x_(i)^(2)=100 then n is

Knowledge Check

  • Consider the Observations x_1=1,x_2=2,x_3=3,.....x_100=100,x_(101)=101,x_(102)=102,x_(103)=103,x_(104)=104 Quarter deviation of the given data is

    A
    26.75
    B
    25
    C
    26
    D
    None of these
  • Consider the Observations x_1=1,x_2=2,x_3=3,.....x_100=100,x_(101)=101,x_(102)=102,x_(103)=103,x_(104)=104 Mean deviation from the median of the given data is

    A
    51/2
    B
    26
    C
    `(51 xx 52)/(103)`
    D
    None of these
  • Consider any set of observations x_(1),x_(2),x_(3),..,x_(101) . It is given that x_(1) lt x_(2) lt x_(3) lt .. Lt x_(100) lt x_(101) , then the mean deviation of this set of observations about a point k is minimum when k equals

    A
    `x_(1)`
    B
    `x_(51)`
    C
    `(x_(1)+x_(2)+..+x_(101))/(101)`
    D
    `x_(50)`
  • Similar Questions

    Explore conceptually related problems

    Consider the algebraic expression (x+1)(x+2)(x+3)(x+4)........(x+100) . Coefficient of x^(99) is equal to

    If average of 20 observations x_(1), x_(2), ........... x_(20) .is y, then the average of x_(1) -101, x_(2) -101, x_(3)-101, .......... X_(20)-10 is:

    If the median of the data , x_(1) , x_(2) , x_(3) , x_(4) , x_(5) , x_(6) , x_(7) , x_(8) is a , then find the median of the data is x_(3) , x_(4) , x_(5) , x_(6) . (where x_(1) lt x_(2) lt x_(3) lt x_(4) lt x_(5) lt x_(6) lt x_(7) lt x_(8) )

    Consider the quantities such that x_(1),x_(2),….x_(10),-1 lex_(1),x_(2)….,x_(10)le 1 and x_(1)^(3)+x_(2)^(3)+…+x_(10)^(3)=0 , then the maximum value of x_(1)+x_(2)+….+x_(10) is

    If average of 20 observation x_(1), x_(2), ……x_(20) is y . then the average of x_(1) - 101, x_(2) - 101, x_(3) - 101, ……. X_(20) - 101 is :