Question

The mean of four positive integers is 5


The median of the four integers is 6



What is the mean of the largest and smallest of the integers?

Answer 1

Given :  The mean of four positive integers is 5 ,  The median of the four integers is 6

To Find : What is the mean of the largest and smallest of the integers?

Solution:

Let say four numbers  are

A , B  , C , D

arranged in ascending order

A - Smallest

D - Largest

Mean = (A + B + C + D)/4

=> 5 = (A + B + C + D)/4

=> A + B + C + D = 20

Median = ( B + C)/2

=> 6 = (B + C)/2

=> B + C  = 12

=> A + 12 + D  =  20

=> A + D  = 8

=> (A + D)/2 = 4

mean of the largest and smallest of the integers = 4

Few example  Satisfying all these : ( for understanding)

1  6  6  7  

1  5  7   7

Learn More:

The mean and median of the data x+a, x+b, x+c are equal and a less ...

https://brainly.in/question/2870143

Find the mean and median of 10 47 3 9 17 4 27 48 12 15 - Brainly.in

https://brainly.in/question/8882216

Answer 2

Answer:

4

Step-by-step explanation:

     Let those positive integers are x₁, x₂, x₃ and x₄ (in ascending order).

Given,  mean = 5

⇒ (x₁ + x₂ + x₃ + x₄)/4 = 5

⇒ x₁ + x₂ + x₃ + x₄ = 20     ...(1)

     Acc. of question, median = 6

                   ⇒ (x₂ + x₃)/2 = 6

In (1),

⇒ x₁ + x₂ + x₃ + x₄ = 20

⇒ x₁ + x₄ + 2(x₂ + x₃)/2 = 20

⇒ x₁ + x₄ + 2(6) = 20  

⇒ x₁ + x₄ = 8

∴ Mean of smallest and largest = mean of extremes

             ⇒ (x₁ + x₄)/2

             ⇒ 8/2

              ⇒ 4

Used: 1,3,5,7,9,11

 Mean = sum/no.    *of observations

 Median = (n+1)/2 th term        [if n is odd]

 median = [n/2 th term + (n/2 + 1)th term]/2   [if n is odd]