91Ó°ÊÓ

The normal distribution can often be used to approximate the binomial distribution, simplifying calculations especially for large sample sizes. But how do you know if it's appropriate to use this approximation?

The rule of thumb is to check whether both \(np\) and \(n(1-p)\) are greater than 5. These conditions ensure the binomial distribution looks enough like a normal distribution. In our example:
\ \

• \(np = 75 \times 0.75 = 56.25\)


• \(n(1-p) = 75 \times 0.25 = 18.75\)

Both values are clearly greater than 5, confirming we can proceed with the normal approximation.

Using the normal distribution approximation helps to streamline calculations, particularly when dealing with large numbers, as it avoids the cumbersome binomial coefficient computation.

The binomial distribution is used to model the number of successes in a fixed number of trials, each with the same probability of success. Here are the key conditions:

• Fixed Number of Trials (\(n\)): The process must have a set number of trials. In the exercise, \(n = 75\).


• Two Possible Outcomes: Each trial has only two outcomes, success (\(p\)) or failure (\(1-p\)). Here, \(p = 0.75\).


• Independent Trials: The result of one trial does not affect another. This is implied but often needs confirmation in real-world problems.


• Constant Probability: The probability of success remains the same for each trial. In this task, \(p = 0.75\) for all 75 trials.

When these conditions are satisfied, you can apply the binomial probability formula: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] This formula calculates the probability of a specific number of successes (\(x\)) occurring out of \(n\) trials.

The z-score is a measure that describes a value's relationship to the mean of a group of values. It is expressed in terms of standard deviations from the mean. Here's how to calculate it:

In our problem, the mean (μ) and standard deviation (σ) are:

• Mean (\( μ = np = 56.25 \))
• Standard Deviation (\( σ = \sqrt{np(1-p)} = 3.75 \))

To find the z-score for \(x = 60\):
\[ z = \frac{x - μ}{σ} = \frac{60 - 56.25}{3.75} = 1 \] This z-score indicates that 60 successes is one standard deviation above the mean.

With this z-score, you can then use the standard normal distribution table to find the corresponding probability. For a z-score of 1, the cumulative probability is approximately 0.8413.

Remember to apply continuity correction when using the normal approximation for discrete data. For \(X = 60\), consider the interval \(P(59.5 < X < 60.5)\) to get a more accurate result.