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Eddi Din [679]
3 years ago
14

Suppose we roll a fair six-sided die and sum the values obtained on each roll, stopping once our sum exceeds 307. Approximate th

e probability that at least 81 rolls are needed to get this sum
Mathematics
1 answer:
SOVA2 [1]3 years ago
5 0

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

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.

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