If 10 fair dice are rolled, approximate the probability that the sum of the values obtained ( which ranges from 10 to 60 is between 30 and 40 inclusive)

Let x be the number showing on the face of the dice

Let p(x) be probability of the face showing = 1/6

```
# A sanity check in R on dice throws
px=1/6
ind=seq(6)
mu=0
for (i in ind) {mu=mu+i*px}
cat('mean is',mu,'\n')
v = 0
for (i in ind){v = v+i*px*(i-mu)^2}
cat('variance is ',v,'\n')
```

My text book shows the following problem.

If 10 fair dice are rolled, approximate the probability that the sum

of the values obtained ( which ranges from 10 to 60 is between 30 and

40 inclusive)

The worked answer shows

Let X_i be the number showing on the i^{th} dice

Let

E[X_i] =\sum_1^{6} p(x_i)x_i = \frac{7}{2}=3.5

Let

Var[X_i]= \sum_1^{6} \frac{(x_i-3.5)^2}{6} = \frac{35}{12}

Let

X = \sum_1^{10} Xi

Then the expected value of X is given by

E[X]= \sum_1^{10} E(X_i)=10\left(\frac{7}{2}\right) = 35

Since, X_i’ s are independent and identically distributed, we have

Var[X]= \sum_1^{10}Var[X_i] =10\left(\frac{35}{12}\right) = \frac{350}{12}

Now we want the probability that the sum of the 10 values is between 30 and 40

By central limit theorem, X \sim \mathcal{N}(E[X],Var[X])

by continuity correction

P(30 \leq X \leq 40) = P(29.5 \leq X \leq 40.5)

using the general conversion formula

Z=\frac{X-\mu_{X}}{\sigma_{X}}

where \sigma_{x_i}= \sqrt(350/12) we can write

P \left( \frac{29.5-35}{\sqrt(\frac{350}{12})} \leq \frac{X-35}{\sqrt(\frac{350}{12})} \leq \frac{40.5-35}{\sqrt(\frac{350}{12})} \right)

P \left( -1.01840 <Z < 1.01840 \right)

where

\phi(Z)=pnorm(1.01840)

and

1-2\phi(Z) = 0.6915

```
# R crosscheck
a1=(29.5-35)/(sqrt(350/12))
a2=(40.5-35)/(sqrt(350/12))
a3 = 1 -2*pnorm(a1)
```