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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 pictureenter image description here

However, the answer in the textbook is bigger than 0.8

Learner132
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2 Answers2

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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
rbm
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    +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 https://stats.stackexchange.com/a/116913/919) of $0.930347...$. Thus the continuity correction supplies more than two more significant figures of accuracy. – whuber Apr 13 '17 at 14:49
  • @rbm: In `NORMDIST`, why have we used fourth parameter as `TRUE` though the `x` takes discrete values? – Shridhar R Kulkarni Nov 16 '17 at 03:22
  • 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 Nov 16 '17 at 09:09
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in Excel

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