Question

Choose the best answer. Let X represent the outcome when a fair six-sided die is rolled. For this random variable,

μX=3.5 and σX =1.71.
If this die is rolled 100 times, what is the approximate probability that the total score is at least 375? (a) 0.0000 (b) 0.0017 (c) 0.0721 (d) 0.4420 (e) 0.9279

Answer 1

Using the Central Limit Theorem, the expected total when rolling the die 100 times is 350 with a standard deviation of 17.1. A z-score calculation gives us the number of standard deviations the score of 375 is above the mean. The probability that the total score is at least 375 is approximately 0.0721.

Step-by-step explanation:

To solve this problem, we can use the Central Limit Theorem (CLT), which states that when an independent random variable is sampled repeatedly, the mean of the sample will form a normal distribution (a bell curve) if the sample size is sufficiently large. In our case, X represents the outcome of a single die roll, and we are given that the mean (μX) is 3.5 and the standard deviation (σX) is 1.71.

When the die is rolled 100 times, the expected total sum would be 100 times the mean of a single die roll, so:

• Expected total sum (μTotal) = 100 * μX = 100 * 3.5 = 350.

Similarly, the standard deviation of the total sum (σTotal) is the standard deviation of a single die roll multiplied by the square root of the number of dice:

• Standard deviation of total sum (σTotal) = √100 * σX = 10 * 1.71 = 17.1.

To find the probability that the total score is at least 375, we can use the normal distribution and z-score. The z-score is calculated by taking the difference between the target total score and the expected total sum, and then dividing by the standard deviation of the total sum:

• Z = (Target total score - μTotal) / σTotal = (375 - 350) / 17.1 ≈ 1.46.

The z-score tells us how many standard deviations away from the mean our target value is. Consulting a z-table or using a calculator would give us the probability of having a sum less than the target. To find the probability of at least the target, we subtract that value from 1.

For example, if the probability of having a sum less than 375 is 0.9279, then the probability of having a sum of at least 375 is 1 - 0.9279 = 0.0721.

The correct answer would therefore be (c) 0.0721, given our approximation.

Answer 2

The approximate probability that the total score of 100 rolls of a fair die is at least 375 is 0.0721, calculated using the central limit theorem and standard normal distribution.

Step-by-step explanation:

To solve this problem, we use the central limit theorem which implies that when a fair die is rolled a large number of times, the sum of the dice rolls will approximate a normal distribution. Given that the expected or mean outcome of a single roll (\(\mu_X\)) is 3.5 and the standard deviation (\(\sigma_X\)) is 1.71, we can calculate the mean and standard deviation for 100 rolls. The mean for 100 rolls will be \(100 \times \mu_X = 100 \times 3.5 = 350\) and the standard deviation will be \(\sigma_{100X} = \sqrt{100} \times \sigma_X = 10 \times 1.71 = 17.1\).

We are interested in finding the approximate probability that the total score is at least 375 when rolling the die 100 times. To find this, we can use a Z-score, which is calculated by \(Z = \frac{X - \mu}{\sigma}\). In our case, we want to know the probability for \(X \geq 375\), which means we calculate the Z-score using \(Z = \frac{375 - 350}{17.1}\).

Plugging 375 into the formula, we get \(Z = \frac{375 - 350}{17.1} \approx 1.46\). Using a standard normal distribution table or a calculator, we can find the probability corresponding to a Z-score of 1.46. The area to the left of Z = 1.46 is approximately 0.9279. Since we want the probability to the right, we subtract this from 1, yielding an approximate probability of \(1 - 0.9279 = 0.0721\), which is option (c).