Final answer:
The question is about finding the probability of rolling fewer than 15 twos (fewer than 30% of 50 rolls) on a modified six-sided die using the normal approximation to the binomial distribution. We calculate the mean and standard deviation for the binomial distribution, find the z-score for 15, and use the standard normal distribution table to find the required probability.
Step-by-step explanation:
The question asks for the probability that fewer than 30% of the rolls result in a two when rolling a fair six-sided number cube with faces 1, 1, 2, 2, 5, 6 fifty times. To solve this, we treat the number of times a two appears as a binomial random variable with n = 50 trials, each with a success probability p = 2/6 (since there are two faces with the number two out of six faces). Fewer than 30% of the rolls is fewer than 15 rolls since 30% of 50 is 15.
Conventionally, one would use a binomial distribution to determine this probability. However, because n is large and p is not extreme, we can use the normal approximation to the binomial distribution. First, we calculate the mean (μ) and standard deviation (σ) for the distribution. The mean is μ = np and the standard deviation is σ = √(np(1-p)). Plugging in our values, we get:
- μ = 50 * (2/6)
- σ = √(50 * (2/6) * (4/6))
Next, we find the z-score for 15 successes (which is 30% of 50), which is calculated using the formula z = (X - μ) / σ, where X is 15 in this case. The resulting z-score is then used to find the probability in a standard normal distribution table. The probability from the table gives us the likelihood that we observe fewer than 15 twos.