Final answer:
The student is asking for an approximation of the probability P(36≤Y≤48) for a random sample of size 12 from a uniform distribution. We use the normal approximation to the binomial by finding the distribution's mean and variance, applying a continuity correction, and then calculating the Z-scores to use with the standard normal distribution.
Step-by-step explanation:
The student is asking for an approximate value of the probability P(36≤Y≤48) for the sum Y of a random sample of size 12 from a distribution with a probability mass function (pmf) that is uniform over the integers 1 to 6. Since the sum of a large number of random variables from a distribution can be approximated by a normal distribution (Central Limit Theorem), we will use the normal approximation to the binomial distribution.
First, find the mean (μ) and variance (σ2) of the given distribution. Since the distribution is uniform over integers 1 through 6, μ = (1+2+3+4+5+6)/6 = 3.5. The variance for a single random variable X from this distribution is σ2 = ∑(x-μ)2*p(x), with each outcome being equally likely (probability = 1/6).
The variance σ2 of a single trial is (1/6)*((1-3.5)2 + (2-3.5)2 + ... + (6-3.5)2) = 2.92. For a sample size of 12, the mean of Y is then 12*μ = 42, and the variance of Y is 12*σ2 = 35.04, and hence the standard deviation is σ = √35.04.
Then, apply the continuity correction by rewriting the probability as P(35.5 < Y < 48.5) to make the discrete problem continuous. Use the normal distribution to approximate the probability by converting Y to its corresponding Z-scores and using standard normal tables or a calculator equipped with normal distribution functions.