Lily asked in Science & MathematicsMathematics · 10 months ago

# Can you find a 5 numbers with a mean of 6 , a median of 7 and a mode of 8 ?

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• geezer
Lv 7
10 months ago

5 numbers ... mean of 6 ... the total of the numbers added together must be 30

5 numbers ... median of 7 ... the number in the middle must be 7

5 numbers ... mode of 8 ... so 8 must be used the most number of times

So ..

If there are 5 numbers and we know it's xx7xx then the 8 is there twice .. xx788

7 + 8 + 8 = 23 ... so the first 2 numbers add up to 7.

So .. the number could be 16788 .. 25788 .. 34788

• 10 months ago

8, 6, 8, 1, 7

2, 8, 7, 5, 8

3, 4, 8, 7, 8

• 10 months ago

3,4,7,8,8

fits the description

Lv 7
10 months ago

There are five numbers and the median is 7, so 7 is the middle (third) number.

There have to be more 8s than any other number, so the fourth and fifth numbers must be 8.

The sum of all the numbers must be 5*6=30.

Sum of first number and second number = 30 - 7 - 8 - 8 = 7.

Since the mode is 8, the first and second numbers must be different and neither can be 7.

There is more than one solution.

{1,6,7,8,8}, {2,5,7,8,8}, {3,4,7,8,8}

• 10 months ago

The median of 5 numbers will be the third number when sorted.  We know that's a 7 so from that alone we know this:

a b 7 x y

Two numbers smaller and two numbers larger than 7.

We are told the mode is 8, which means there are at least 2 8's in the set and it has the most numbers in the set.  There is room for 2 numbers larger than 7's, so these are both 8s.  Since there are two of them, no other number can have more than 1 for 8 to be the only mode:

a b 7 8 8

we now need a mean of 6.  There are an infinite number of solutions here, but presuming you want only integers, we can figure this out.

The mean is the sum of the data point values divided by the number of data points.  Do this and set it equal to 6 and we can come up with an expression that compares a and b:

(a + b + 7 + 8 + 8) / 5 = 6

(a + b + 23) / 5 = 6

a + b + 23 = 30

a + b = 7

So we know that the sum of a and b is 7, and both must be less than 7 and not the same.  Keeping these as integers, we don't have them completely narrowed down to one possible set, but a small set of sets:

The solution can be either:

1 6 7 8 8

2 5 7 8 8

3 4 7 8 8

All three of these sets satisfy the conditions of your set.  We would need more information in order to narrow it down further.

• 10 months ago

a , b , c , d , e

These are sorted from least to greatest.  We know that c = 7

a , b , 7 , d , e

We know that the most common number is 8, so d and e are 8

a , b , 7 , 8 , 8

We know that a and b can't be equal to each other, or else 8 wouldn't be the mode

a < b

We know that the mean is 5

5 = (a + b + 7 + 8 + 8) / 5

25 = a + b + 23

2 = a + b

As long as a + b = 2, we're good.  Are there any other restrictions?  Can a be 0?  If so, we have:

0 , 2 , 7 , 8 , 8

That fits everything you need.

Do you see how straightforward this stuff is?  There's nothing crazy or ridiculous here.  All you have to do is work through it methodically, piece-by-piece, until you have an answer.

EDIT:

Mean of 6.  Whatever.  Same thing.

6 = (a + b + 7 + 8 + 8) / 5

30 = a + b + 23

7 = a + b

a = 1 , b = 6

a = 2 , b = 5

a = 3 , b = 4

There you go.

1 , 6 , 7 , 8 , 8

2 , 5 , 7 , 8 , 8

3 , 4 , 7 , 8 , 8