Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

Compute the median from the following table

36.55
35.55
40.05
None of these

The correct answer is: 36.55

$n equals sum f equals 159$ . Here n = 159, which is odd
$therefore$ Median number $equals fraction numerator 1 over denominator 2 end fraction left parenthesis n plus 1 right parenthesis equals fraction numerator 1 over denominator 2 end fraction left parenthesis 159 plus 1 right parenthesis equals 80$ ,
which is in the class 30-40. (see the row of cumulative frequency 95, which contains 80).
Hence median class is 30-40.
\ We have,
l = Lower limit of median class = 30
f = Frequency of median class = 45
C = Total of all frequencies preceding median class
= 50
i = Width of class interval of median class = 10
\ Required median $equals l plus fraction numerator fraction numerator N over denominator 2 end fraction minus C over denominator f end fraction cross times i$
$equals 30 plus fraction numerator fraction numerator 159 over denominator 2 end fraction minus 50 over denominator 45 end fraction cross times 10 equals 30 plus fraction numerator 295 over denominator 45 end fraction equals 36.55$ .

Related Questions to study

The following data gives the distribution of height of students

The median of the distribution is[AMU 1994]</span

The following data gives the distribution of height of students

The median of the distribution is[AMU 1994]</span

What is the standard deviation of the following series

What is the standard deviation of the following series

Compound (A) The Compound (A) is:

Compound (A) The Compound (A) is:

chemistry-General

parallel

The order of [x y z] $open square brackets table row a h g row h b f row g f c end table close square brackets open square brackets table row x row y row z end table close square brackets$ is

The order of [x y z] $open square brackets table row a h g row h b f row g f c end table close square brackets open square brackets table row x row y row z end table close square brackets$ is

If the matrix $open square brackets table row 1 3 cell alpha plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then $alpha$ =

If the matrix $open square brackets table row 1 3 cell alpha plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then $alpha$ =

If the matrix $open square brackets table row 1 3 cell lambda plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then the value of  is :

If the matrix $open square brackets table row 1 3 cell lambda plus 2 end cell row 2 4 8 row 3 5 10 end table close square brackets$ is singular, then the value of  is :

parallel

Let A = $open square brackets table row cell negative 1 end cell 2 cell negative 3 end cell row cell negative 2 end cell 0 3 row 3 cell negative 3 end cell 1 end table close square brackets$ be a matrix, then(determinant of A ) × (adjoint of inverse of A)is equal to-

Let A = $open square brackets table row cell negative 1 end cell 2 cell negative 3 end cell row cell negative 2 end cell 0 3 row 3 cell negative 3 end cell 1 end table close square brackets$ be a matrix, then(determinant of A ) × (adjoint of inverse of A)is equal to-

Let X= $open square brackets table row x row y row z end table close square brackets$ , D = $open square brackets table row 3 row 5 row 11 end table close square brackets$ and A = $open square brackets table row 1 cell negative 1 end cell cell negative 2 end cell row 2 1 1 row 4 cell negative 1 end cell cell negative 2 end cell end table close square brackets$ . If X = A^–1 D, then X is equal to

Let X= $open square brackets table row x row y row z end table close square brackets$ , D = $open square brackets table row 3 row 5 row 11 end table close square brackets$ and A = $open square brackets table row 1 cell negative 1 end cell cell negative 2 end cell row 2 1 1 row 4 cell negative 1 end cell cell negative 2 end cell end table close square brackets$ . If X = A^–1 D, then X is equal to

If k . $open square brackets table row cell negative 1 end cell 2 2 row 2 cell negative 1 end cell 2 row 2 2 cell negative 1 end cell end table close square brackets$ is an orthogonal matrix, then k is equal to-

If k . $open square brackets table row cell negative 1 end cell 2 2 row 2 cell negative 1 end cell 2 row 2 2 cell negative 1 end cell end table close square brackets$ is an orthogonal matrix, then k is equal to-

parallel

If A is square matrix such that A² = A, then |A| equals-

If A is square matrix such that A² = A, then |A| equals-

If the matrix A = $open square brackets table row 8 cell negative 6 end cell 2 row cell negative 6 end cell 7 cell negative 4 end cell row 2 cell negative 4 end cell lambda end table close square brackets$ is singular, then $lambda$ equal to:

If the matrix A = $open square brackets table row 8 cell negative 6 end cell 2 row cell negative 6 end cell 7 cell negative 4 end cell row 2 cell negative 4 end cell lambda end table close square brackets$ is singular, then $lambda$ equal to:

If A² = 8A + kI where A = $open square brackets table row 1 0 row cell negative 1 end cell 7 end table close square brackets$ then k is :

If A² = 8A + kI where A = $open square brackets table row 1 0 row cell negative 1 end cell 7 end table close square brackets$ then k is :

parallel

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ and B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ such that A² = B the $alpha$ n is :

If A = $open square brackets table row alpha 0 row 1 1 end table close square brackets$ and B = $open square brackets table row 1 0 row 5 1 end table close square brackets$ such that A² = B the $alpha$ n is :

Matrix M_r is defined as M_r= $open square brackets table row r cell r – 1 end cell row cell r – 1 end cell r end table close square brackets$ , r $element of$ N. Then det. (M₁) + det.(M₂) + ........... + det.(M₂₀₀₉) = ?

Matrix M_r is defined as M_r= $open square brackets table row r cell r – 1 end cell row cell r – 1 end cell r end table close square brackets$ , r $element of$ N. Then det. (M₁) + det.(M₂) + ........... + det.(M₂₀₀₉) = ?

if $open square brackets table row 1 2 3 row 3 1 2 row 2 3 1 end table close square brackets$ $open square brackets table row x row y row z end table close square brackets$ = $open square brackets table row cell 4 – 2 end cell row cell 0 – 6 end cell row cell – 12 end cell end table close square brackets$ $open square brackets table row 2 row 1 end table close square brackets$ then (x, y, z) is equal to -

if $open square brackets table row 1 2 3 row 3 1 2 row 2 3 1 end table close square brackets$ $open square brackets table row x row y row z end table close square brackets$ = $open square brackets table row cell 4 – 2 end cell row cell 0 – 6 end cell row cell – 12 end cell end table close square brackets$ $open square brackets table row 2 row 1 end table close square brackets$ then (x, y, z) is equal to -

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.