Question

Suppose we roll a fair six-sided die and sum the values obtained on each roll, stopping once our sum exceeds 307. Approximate the probability that at least 81 rolls are needed to get this sum

Answer 1

Answer:

There is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

Step-by-step explanation:

The roll of a six-sided die sum has this parameters:

Mean: 3.5

Standard deviation: 1.7

If the die is rolled 81 times, the distribution of the sum of the 81 rolls will have the following parameters:

$\mu=n*M=81*3.5=283.5\\\\\sigma=\sqrt{n}s=\sqrt{81}*1.7=9*1.7=15.3$

Note: the variance of the sum of random variables is equal to the sum of the variance of the individual variables.

Then, we can calculate the probabilties that the sum of 81 rolls is lower than 307.

$z=\frac{x-\mu}{\sigma}=\frac{307-283.5}{15.3}= \frac{23.5}{15.3}= 1.536\\\\\\P(x<307)=P(z<1.536)=0.94$

There is 94% of chances that the sum of 81 rolls is lower than 307, so there is a probability of P=0.94 that at least 81 rolls are needed to get the sum of 307.

Answer 2

The probability will be "0.9370".

According to the question,

• $\sigma = 1.71$
• $\mu = 3.5$

By using the Central limit theorem, we get

→ $S_{81} \sim N(81\times \mu, \sqrt{81\times \sigma^2} )$

        $\sim N (81\times 3.5, 9\times \sigma)$

        $\sim N (283.5, 9\times 1.71)$

        $\sim N (283.5, 15.39)$

hence,

→ $P(S_{81} \leq 307) = P(Z \leq \frac{307-283.5}{15.39} )$

                        $= P(Z \leq 1.53)$

                        $= 0.9370$

Thus the above approach is correct.  

Learn more about probability here:

https://brainly.com/question/14291600