# roll dice 400 times, what is the probability that the total number of dots is at least 1350?

I know how to solve if roll dice 2 time and the sum is 10, but by drawing table. Therefore, with 400 times and sum is 1350, that is ridiculous.

Let's say using Excel to solve the problem.

Should I use Binomial random function BINOM.DIST.RANGE(trials,probability_s,number_s,[number_s2]). But the thing is how to know probability of getting 1350 ?

I also thought about using Normal random variable function NORMDIST(x,mean,standard_dev,cumulative) using the data in the picture

However, the answer in the textbook is bigger than 0.8

• You don't have a binomial distribution in this problem. It's not relevant to this question Commented Apr 14, 2017 at 6:48

I think your problem is that you are mixing variance and standard deviation in the NORM.DIST.

You should use =1-NORMDIST(1350,1400,SQRT(1167),TRUE) in Excel,

or in R:

sims  <- 400
mu    <- 7/2
sigma <- 35/12
pnorm(1350, mean = 400 * mu, sd = sqrt(400 * sigma), lower.tail = F)


giving

[1] 0.9283825

• +1. Note that the usual continuity correction will work noticeably better, even with these seemingly large numbers: pnorm(1350 - 1/2, mean = 400 * mu, sd = sqrt(400 * sigma), lower.tail = F) gives $0.930361$, in agreement to almost five significant figures with the exact value (computed as described at stats.stackexchange.com/a/116913/919) of $0.930347...$. Thus the continuity correction supplies more than two more significant figures of accuracy.
– whuber
Commented Apr 13, 2017 at 14:49
• @rbm: In NORMDIST, why have we used fourth parameter as TRUE though the x takes discrete values? Commented Nov 16, 2017 at 3:22
• The true has nothing to do with discrete/continuous, it is a cumulative flag, i.e. specifying whether you want to calculate PDF or CDF. Consider NORMDIST(0,0,1,FALSE) (which equals ~0.39) and NORMDIST(0,0,1,TRUE) (which is exactly 0.50!).
– rbm
Commented Nov 16, 2017 at 9:09

in Excel

=1-NORMDIST(1350;1400;SQRT((VAR.P(1;2;3;4;5;6)*400));1)