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When we talk about probability in the context of binomial distribution, we're dealing with events that have two possible outcomes: success or failure. In our exercise, a driver wearing a seatbelt is considered a success. The probability of success, denoted as \( p \), is 0.75, indicating that 75% of drivers regularly wear seatbelts. Conversely, the probability of failure is \( 1 - p = 0.25 \).

To calculate probabilities using the binomial distribution, we rely on the binomial probability formula:

• \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

Here, \( n \) is the total number of trials, \( k \) is the number of successful outcomes we are interested in, and \( \binom{n}{k} \) is the binomial coefficient or "n choose k."

In situations where \( n \) is large, calculating probabilities directly using this formula can be cumbersome, which is why normal approximation is so helpful.

Normal approximation is a method used to estimate the binomial distribution's probabilities when the number of trials \( n \) is large. This is because the binomial distribution tends to resemble a normal distribution under such conditions.

In our problem, we have \( n = 500 \), which is sufficiently large to use this approximation. To apply it, we first calculate the mean \( \mu \) and standard deviation \( \sigma \) of the binomial distribution:

• \( \mu = n \times p = 500 \times 0.75 = 375 \)
• \( \sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{500 \times 0.75 \times 0.25} \approx 9.68 \)

Using these, the binomial distribution \( X \sim B(n, p) \) can be approximated by a normal distribution \( X \sim N(\mu, \sigma^2) \). This approximation enables us to leverage the easy-to-use properties of the normal distribution for calculating probabilities.

Z-scores are a powerful tool when working with normal distributions. They help us understand how far a particular value is from the mean, measured in terms of standard deviations. The formula to calculate a Z-score is:

• \( Z = \frac{X - \mu}{\sigma} \)

Where \( X \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

For example, to find the probability that between 360 and 400 drivers wear seatbelts, we calculate the Z-scores for 359.5 (to include all values up to 360 using continuity correction) and 400.5. These calculations give us:

• For 359.5, \( Z \approx -1.60 \)
• For 400.5, \( Z \approx 2.64 \)

We then use Z-tables to find the probabilities corresponding to these Z-scores, helping us determine the portion of this normal distribution that lies between these two points.

Continuity correction is an adjustment made when using normal approximation to the binomial distribution. It's needed because the normal distribution is continuous, whereas the binomial distribution is discrete.

In the exercise, we wanted to find the probability for "between 360 and 400 drivers." We adjusted this to "359.5 to 400.5," adding 0.5 to include the endpoints in a continuous manner. This small shift in values, known as continuity correction, ensures that our normal approximation accurately reflects the discrete nature of real-world data, enabling more precise probability calculations.

• For endpoint adjustments: subtract 0.5 from lower bounds and add 0.5 to upper bounds

By applying this correction, you can align your discrete range with the continuous model, which leads to more accurate results when using the normal distribution methods.