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The Z-score is an essential statistic to understand when dealing with normal distributions. It tells us how many standard deviations an element is from the mean. In this context, the Z-score helps us determine the probability of a particular height among 10-year-olds relative to the average. To calculate a Z-score, use the formula \( Z = \frac{X - \mu}{\sigma} \). Here, \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For instance, if the mean height is 55 inches and the standard deviation is 6 inches, the Z-score for a height of 54 inches would be \( Z = \frac{54 - 55}{6} = -0.1667 \). This negative Z-score indicates that 54 inches is below the average height. Then, using a standard normal distribution table, we find that roughly 43.38% of 10-year-olds fall below this height, meaning they wouldn't be able to go on the ride.

The binomial distribution is crucial when dealing with experiments that have two possible outcomes, such as success or failure. In the context of our problem, success is defined as being tall enough to ride the ride, while failure is too short. A big advantage of the binomial distribution is its ability to calculate the probability of having a certain number of successes in a set of trials. In practice, the probability of being able to ride is 0.5662. If we have four children and want the probability that at least two can ride, we must find the total possibility for zero and one child riding and subtract this from one. Use the formula: \( P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \). After computing, you find roughly 82.83% of scenarios result in at least two children being able to ride.

Geometric probability helps us calculate the chance of the first success occurring on a certain trial. It's particularly useful when you want to know how many trials until your first success happens, which is the essence of Part (c) from the exercise. Here, the probability of failure until the first success is significant. If we need the third child entering to be the first one tall enough to ride, then the first two mustn't be tall enough. The scenario occurs with two initial failures followed by a success: \( P(\text{first success on 3rd trial}) = (0.4338)^2 \times 0.5662 \approx 0.1064 \). This probability indicates the likelihood of witnessing this precise sequence of events.

The negative binomial distribution extends the geometric probability concept to more complex scenarios, like finding the probability of a specific success occurring on a target trial number. In Part (d), we are tasked with identifying the chance that the fifth child tallying up to success is the twelfth to enter the park. Here, the calculation reflects the probability of achieving exactly four successes in 11 trials before the fifth one. The equation for this probability takes the form:\[ P = \binom{11}{4} \times (0.5662)^5 \times (0.4338)^7 \] Compute this to find the probability of approximately 11.33%, which describes the likelihood of the fifth person tall enough being the twelfth customer.