Chapter 8: Q 8.6 (page 390)

A die is continually rolled until the total sum of all rolls exceeds 300. Approximate the probability that at least 80 rolls are necessary.

Short Answer

Expert verified

$P {Y_{79} \leq 300} = 0.9429$

Step by step solution

Step 1. Given information

Let X_idenote the value of the die, of i^throll.

Step 2. Mean and variance of Xi

Mean = $E (X_{i}) = \frac{(1 + 2 + 3 + 4 + 5 + 6)}{6} = \frac{21}{6} = \frac{7}{2}$

Variance =role="math" localid="1648052296820" $V (X_{i}) = \frac{35}{12}$

Step 3. Introducing Yn

Assuming that die is continually rolled until the total sum of all rolls exceeds 300.

Let Y_nrepresents the total sum of all rolls with n number of rolls.

$Y_{n} = {\sum X_{i}}_{1}^{n}$

Step 4. Mean and variance of Yn

Mean = $E (Y_{n}) = n \times E (X_{i}) = \frac{7 n}{2}$

Variance = $V (Y_{n}) = n \times V (X_{i}) = \frac{35 n}{12}$

Step 5. To find

The probability that at least 80 rolls are necessary can be written as -

$P \{{\sum X_{i} \leq 300}_{1}^{79}\} = P \{Y_{79} \leq 300\}$

Step 6. Using central limit theorem

$P {Y_{79} \leq 300} = P \{\frac{Y_{79} - \frac{7 (79)}{2}}{\sqrt{\frac{35 (79)}{12}}} \leq \frac{300 - \frac{7 (79)}{2}}{\sqrt{\frac{35 (79)}{12}}}\} P {Y_{79} \leq 300} = P \{Z \leq 1.58\} P {Y_{79} \leq 300} = 0.9429$

Step 7. Final answer

The probability that at least 80 rolls are necessary is 0.9429

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A die is continually rolled until the total sum of all rolls exceeds 300. Approximate the probability that at least 80 rolls are necessary.

Short Answer

Step by step solution

Step 1. Given information

Step 2. Mean and variance of Xi

Step 3. Introducing Yn

Step 4. Mean and variance of Yn

Step 5. To find

Step 6. Using central limit theorem

Step 7. Final answer

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