An orchestra consists of 25 members. The youngest member is 32 years old. She leaves and is replaced by a new member who is 30 years old.whaat is effect on average age , mean??
\[\huge\frac{ 32+x_2+....+x25 }{ 25 }=y\]
previously the average age is (x+32)/25..where x is the sum of ages of other 24 members..now i think you can tell what is the average age after 32 year member is replaced by 30 year old member..
\[\huge \frac{ 30+x_2+.....+x25 }{ 25 }=w\]
\[\huge \frac{25w+2}{ 25 }=y\]
\[\huge w=\frac{ 25y-2 }{ 25 }\]
the mean of their ages will decrease by 0.08 ??
median??
If A is the average age of the 25 people before the 32-year old left, then 25*A is the sum of the ages. After the 32-year old left and a 30-year old joined, then (25*A - 2)/25 is the new average age. (25*A - 2)/25 = 25*A/25 - 2/25 = A - 2/25 = A - 8/100 = A - .08. So, does the average go down by .08 people? If so, I agree with @deena123
Let x = original average age. Original sum of ages = 25x New sum of ages = 25x - 2 New mean is \[\frac{25x-2}{25}=x-0.08\]
> median?? @non008 Are you asking a second question?
Tell me about median ?
no median of above question.
I did the average problem on a smaller case with 3 people and the same problem constraints as posted. The mean went down by 2/3 a point. So, I'm conjecturing that under the same conditions but with 100 people in the group, the mean would drop by 2/100 a unit. Just saying.
The median will not change when the age of the youngest player is reduced. The median is the middle value in a ranked set of data values.
thanks @Directrix and @kropot72
You're welcome :)
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