Answer:
The probability that the sum of the dice is less than 400 is 0.3050.
Step-by-step explanation:
Let X = outcomes of a single roll of a dice.
The possible outcomes of X are
S = {1, 2, 3, 4, 5, 6}
The probability of the random variable X is:
[tex]P(X)=p=\frac{1}{6}=0.167[/tex]
Compute the mean and variance of the random variable X as follows:
[tex]E(X)=\sum x.P(X=x)=3.5\\V(X)=E(X^{2})-(E(X))^{2}=2.917[/tex]
The dice was rolled n = 117 times.
The sum of the values of X in these 117 rolls follows a Normal distribution with mean 3.5 and variance 2.917.
Compute the probability that the sum of the dice is less than 400 as follows:
[tex]P(\sum X<400)=P(\frac{\sum X}{n}<\frac{400}{117})\\=P(\bar X<3.42)\\=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{3.42-3.5}{\sqrt{2.917/117}})\\=P(Z<-0.51)\\=1-P(Z<0.51)\\=1-0.695\\=0.305[/tex]
*Use the z-table for the probability.
Thus, the probability that the sum of the dice is less than 400 is 0.3050.
