Question

A 6-sided die is repeatedly rolled until the total sum of all the rolls exceeds 300. Approximate(using the Central Limit Theorem) the probability thatat least80 rolls are necessary to reach asum that exceeds 300.

Answer 1

0.9430

Explanation:

The other answer is almost right but answers the wrong question. The question asks for the probability that at least 80 rolls are necessary not more than 80. So we can use the same process but use where we are using the sum of 79 die. The other answer also misses continuity correction. We must use continuity correction since we're approximating a discrete variable with a continuous one through CLT so P(X <= 300) -> P(X < 300.5). With only these changes, we get 0.943.

Answer 2

There is a probability of 90.49% of needing more than 80 rolls to reach 300.

Explanation:

The Central Limit Theorem tells us that the sampling distribution, as the size of the sample gets larger, approaches the normal distribution. This is independent of the population distribution of the random variable.

Then we can use the normal distribution parameters to model this problem.

Let Y be the sum of 80 dices.

`Y=\sum_(i=1)^(80)X_i\\\\E(Y)=80*E(X_i)=80*[(1/6)*(1+2+3+4+5+6)]=80*3.5=280\\\\V(Y)=80V(X_i)\\\\V(Y)=80*(1/6)[(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2]\\\\V(Y)=80*(1/6)*17.5=233.33`

`\sigma=√(V(Y))=√(233.33)=15.28`

We can use the z-value to calculate the probability of getting 300 or more in 80 rolls.

`z=(X-\mu)/\sigma=(300-280)/15.28=1.31\\\\\\P(X>300)=P(z>1.31)=0.0951\\\\P(X<300)=1-0.0951=0.9049`

There is a probabilitity P=0.9049 that the 80 rolls will not reach a 300 score. Thus we can conclude that there is a probability of 90.49% of needing more than 80 rolls to reach 300.