the average weight of 6 people is 173, a 7th person gets on and now the average weight is 165, how much did the 7th person weigh

For problems like this involving an average of numbers that are "close" together, here is a different strategy for solving the problem that you might like to try.

The original 6 people have an average weight of 173, which is 8 pounds more than the average of all 7 people. So all together, the weight of those 6 people is 6*8=48 pounds over the 165 average. That means the 7th person's weight must be 48 pounds below the 165 average.

So the weight of the 7th person is 165-48 = 117 pounds.

elcharly64 elcharly64 · Answer 2 · 2020-10-07T01:15:46+02:00

Answer:

The 7th person weighted 117

Step-by-step explanation:

Average Value

The mean or average of a number n of measurements is defined as the sum of all values divided by n.

[tex]\displaystyle \bar x=\frac{\sum x_i}{n}\\[/tex]

The question gives us some input data. The average weight of n=6 people is [tex]\bar x=173[/tex]. With this information, we can set this relationship:

[tex]\displaystyle 173=\frac{\sum x_i}{6}[/tex]

From the above equation, we know the sum of the first 6 persons:

[tex]\sum x_i=6*173=1,038[/tex]

When a 7th person gets on, the new average gets down to 165 and n=7. The equation for the new average is:

[tex]\displaystyle 165=\frac{\sum x_i+X}{7}[/tex]

Note we have a new summand to the numerator. It's the weight of the 7th person. Let's solve for X:

[tex]\displaystyle 165=\frac{1,038+X}{7}[/tex]

Multiply by 7:

[tex]7*165=1,038+X[/tex]

Operate and swap sides:

[tex]1,038 +X=1,155[/tex]

[tex]X=1,155-1,038=117[/tex]

The 7th person weighted 117