Chi-squared distribution and Central Limit Theorem approximation I think $m$'s mean is $5$, variance is $2\cdot5=10$
then, $(3.5-5)/\sqrt{\frac{10}{100}}$  it is the lower bound
$(7.1-5)/\sqrt{\frac{10}{100}}$  it is the upper bound
Is this right? But I found the bounds is too large. I even can not find the probability....
what is wrong?
 A: With df = 5, each individual observation $X_i \sim$ Chisq(df=5), for $i = 1,\dots,100,$ does have mean 5 and variance 10 (as you speculate).
Then the  sample mean of 100 has $E(\bar X) = 5$ and $V(\bar X) = 10/100 = 0.1$ or
SD$(\bar X) = \sqrt{0.1} =  0.3162$.
Also,
$$P\{3.5 < \bar X < 7.1\} = P\{(3.5 - 5)/.3162 < Z < (7.1 - 5)/.3162\} = P\{-4.744 < Z < 6.641\} \approx 1,$$
where $\bar X$ is the mean of 100 observations from Chisq(df=5) and $Z$ is standard normal. 
Now you would ordinarily need to use printed tables of the standard normal distribution or statistical software to get the probability. However, almost
all of the probability under a standard normal curve lies between -3 and +3.
These bounds are so far below -3 and so far above +3 that you won't find them
in normal tables. I put $\approx 1$ at the end of the display above to indicate this answer.
I simulated the problem and made a histogram of the means from 100,000 samples of size 100, which was very nearly normal (and with mean almost exactly 10 and SD almost exactly 0.316, as anticipated.)
