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Normal approximation can be used to estimate probabilities in a binomial distribution when certain conditions are met. For a binomial distribution with parameters \( n \) and \( p \), the normal distribution can be a suitable approximation if both \( np \geq 5 \) and \( n(1-p) \geq 5 \). This rule helps ensure that the binomial distribution is roughly symmetric and not too skewed.

When these criteria are satisfied, we can approximate the binomial distribution using a normal distribution with the same mean (\( \mu \)) and standard deviation (\( \sigma \)). The mean of the binomial distribution is \( \mu = np \), and the standard deviation is \( \sigma = \sqrt{np(1-p)} \).

For our example, given \( n=60 \) and \( p=0.4 \), we found that \( np = 24 \) and \( n(1-p) = 36 \), both of which are greater than 5. Hence, we can use the normal approximation. This can simplify large binomial probability calculations that may otherwise be complex.

The binomial coefficient, represented as \( \binom{n}{x} \), is a key component of the binomial probability formula. It tells us the number of ways to choose \( x \) successes out of \( n \) trials regardless of the order. The binomial coefficient is calculated using the formula: \[ \binom{n}{x} = \frac{n!}{x!(n-x)!} \] where \( n! \) is the factorial of \( n \).

In our given problem with \( n=60 \) and \( X=20 \), we calculate \( \binom{60}{20} \) to find the number of ways to choose 20 successes out of 60 trials. Factorials grow rapidly, so it's often easier to use a calculator or software to compute large binomial coefficients.

The z-score is a measure that describes the position of a raw score in terms of its distance from the mean (\( \mu \)) of a distribution, measured in standard deviations (\( \sigma \)). It is calculated using the formula: \[ z = \frac{X - \mu}{\sigma} \]

In the context of normal approximation for a binomial distribution, once we have the mean and standard deviation, we convert the number of successes (\( X \)) to a z-score to find its position on the standard normal distribution.

For our problem where \( X=20 \), \( \mu=24 \), and \( \sigma=3.79 \), the z-score is: \[ z = \frac{20 - 24}{3.79} \approx -1.06 \]

This z-score helps us find the probability associated with \( X=20 \) by looking up \( z=-1.06 \) in the standard normal distribution table. This gives us the approximate probability \( P(Z < -1.06) \approx 0.1446 \).