Using both the binomial distribution and the normal approximation. A die is thrown 720 times. (a) Find the probability that 3 comes up exactly 125 times. (b) Find the probability that 3 comes up between 115 and 130 times.

When dealing with a large number of trials in a binomial distribution, the calculations can get tricky. To make things easier, we use the Normal Approximation. This method transforms our binomial distribution into a normal distribution, which is much simpler to work with when calculating probabilities. This approximation works well when the number of trials, n, is large and the probability, p, is not too close to 0 or 1. In our example, the die is thrown 720 times, and we want to know the probability of rolling a 3 exactly 125 times. By using the normal approximation, we can simplify this problem significantly.

We start by finding the mean, \(\text{mean}\), which is given by the formula \[ \mu = np \]. In this case, \[ \mu = 720 \times \frac{1}{6} = 120 \]. Next, we calculate the standard deviation, \(\text{standard deviation}\), using the formula \[ \sigma = \sqrt{np(1 - p)} \]. For our problem, \[ \sigma = \sqrt{720 \times \frac{1}{6} \times \frac{5}{6}} \approx 10 \]. Using these two values, we can move on to the next steps in our approximation.

The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. To use the normal approximation, we need to transform our binomial variable, X, into a standard normal variable, Z. This process is called standardizing and involves the formula \[ Z = \frac{X - \mu}{\sigma} \]. This transformation helps us use the Z-table, a tool that lists the probabilities associated with standard normal variables.

For instance, to find the probability that we roll a 3 exactly 125 times, we calculate the Z-score using \[ Z = \frac{125 - 120}{10} = 0.5 \]. We then use the Z-table to find the corresponding probability. The Z-table tells us how likely it is for a standard normal variable to fall within a certain range of values. This step is crucial for determining the probability of specific outcomes in our problem.

Calculating the probability involves finding specific values from the standard normal distribution. For our exercise, we want the probability of exactly 125 rolls of 3. After standardizing, we find that \[ P(X = 125) \] becomes \[ P(0.45 < Z < 0.55) \] using the continuity correction (explained in the next section). The Z-table tells us that this probability is approximately 0.08.

Next, we want to calculate the probability of rolling a 3 between 115 and 130 times. First, we standardize these values: \[ Z_{lower} = \frac{115 - 120}{10} = -0.5 \] and \[ Z_{upper} = \frac{130 - 120}{10} = 1 \]. We then use the Z-table to find \[ P(-0.5 < Z < 1) \]. These calculations allow us to determine the likelihood of a 3 appearing between 115 and 130 times.

The continuity correction is a technique used when a discrete binomial distribution is approximated by a continuous normal distribution. Since binomial variables are discrete (specific values), but normal variables are continuous (any value within a range), we adjust our calculations slightly. This correction improves the accuracy of our approximation.

For example, when finding \[ P(X = 125) \], we actually look for the range \[ P(124.5 < X < 125.5) \] instead. This accounts for the continuous nature of the normal distribution. Similarly, when estimating the probability between two values (e.g., students rolling a 3 between 115 and 130 times), we adjust the boundaries: \[ P(114.5 < X < 130.5) \]. This simple adjustment ensures our approximation is as accurate as possible, making our calculations more reliable.

Using these techniques, we can confidently answer questions about binomial probabilities with large trials, turning complex problems into manageable ones.