91Ó°ÊÓ

In statistics, the binomial distribution is used for scenarios where there are a fixed number of trials, each with two possible outcomes: success or failure. But what if your problem involves many trials? That's where the normal approximation comes in handy. When the number of trials is large enough, the binomial distribution can be approximated by the normal distribution.
This means we can use the normal distribution (a bell-shaped curve) to estimate probabilities, making complex calculations simpler. However, certain conditions need to be met for this approximation to be valid. Specifically, both of these conditions must hold:

• \( np \geq 5 \)
• \( n(1-p) \geq 5 \)

These conditions make sure we have a sufficiently large number of trials. If these conditions are satisfied, you can turn your binomial distribution problem into a normal distribution problem, easing the computation of probabilities.

When approximating a discrete binomial distribution with a continuous normal distribution, we apply continuity correction. This step adds half a unit (or 0.5) to the discrete value to get a better fit.
This correction is crucial because the binomial distribution only takes integer values while the normal distribution can take any real number. For our problem with \( x = 18 \), we adjust it to range between \( 17.5 \) and \( 18.5 \).
Continuity correction ensures that your normal approximation aligns more closely with the actual binomial probabilities. It might seem like a small tweak, but it significantly improves accuracy.

Z-scores are a great tool for converting binomial data into a form compatible with the normal distribution. A Z-score measures how many standard deviations an element is from the mean. The formula for calculating Z-scores is:
\[{Z = \frac{X - \mu}{\sigma}}\]
Here, \( X \) is the value you are examining, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For our problem, we calculate Z-scores for both \( 17.5 \) and \( 18.5 \) to account for the continuity correction.
Once you have the Z-scores, you can use Z-tables or statistical calculators to find the probability associated with those scores. This allows you to approximate the binomial probability using the normal distribution accurately.

The mean and standard deviation are the backbone of the normal distribution. For a binomial distribution, the mean (\(\mu\)) and standard deviation (\(\sigma\)) are calculated as follows:

• Mean \(\mu = np \)
• Standard Deviation \(\sigma = \sqrt{np(1-p)}\)

In our exercise, \( n = 80 \) and \( p = 0.15 \). This gives:
Mean: \(\mu = 80 * 0.15 = 12\)
Standard Deviation: \(\sigma = \sqrt{80 * 0.15 * 0.85} \approx 3.177\)
Having these values helps in accurately approximating the binomial distribution as a normal distribution, and they are essential for calculating Z-scores and ultimately determining the necessary probabilities.