Math Help - arithmetic mean

1. arithmetic mean

the arithmetic mean of a set of 50 numbers is 38. if two numbers 45 and 55 are discarded the mean of the remaining set of numbers is ____ .

need little help on this guys.

thanks

3. Re: arithmetic mean

Originally Posted by rcs
the arithmetic mean of a set of 50 numbers is 38. if two numbers 45 and 55 are discarded the mean of the remaining set of numbers is ____ .

need little help on this guys.

thanks
You are told \displaystyle \begin{align*} \frac{t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49}}{50} = 38 \implies t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49} = 1900 \end{align*}. Two of these terms are 45 and 55, so removing them gives

\displaystyle \begin{align*} t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49} - 45 - 55 &= 1900 - 45 - 55 \\ &= 1800 \end{align*}

There are now 48 terms, so the mean of these 48 terms will be \displaystyle \begin{align*} \frac{1800}{48} = 37.5 \end{align*}.

4. Re: arithmetic mean

thanks sir... but how did you get 1900?

5. Re: arithmetic mean

Originally Posted by rcs
thanks sir... but how did you get 1900?
$50 \times 38 = 1900$

6. Re: arithmetic mean

If the arithmetic mean is 38, then their sum is 50*38 = 1900...

7. Re: arithmetic mean

you are all God's gifts to all not so good in mathematics... God Bless guys

8. Re: arithmetic mean

Originally Posted by Prove It
You are told \displaystyle \begin{align*} \frac{t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49}}{50} = 38 \implies t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49} = 1900 \end{align*}. Two of these terms are 45 and 55, so removing them gives

\displaystyle \begin{align*} t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49} - 45 - 55 &= 1900 - 45 - 55 \\ &= 1800 \end{align*}

There are now 48 terms, so the mean of these 48 terms will be \displaystyle \begin{align*} \frac{1800}{48} = 37.5 \end{align*}.
Thank you King of Proofs ProveIt