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The binomial distribution is a common probability distribution that models the number of successes in a fixed number of trials of a binary experiment. In simpler terms, it helps find the likelihood of something happening a certain number of times out of a set number of events. Consider flipping a coin a specific number of times, where you count the number of heads that appear. Here, each flip is an independent trial.

When solving the given problem about social networking use among young adults, we treat each individual sampled as a trial. A "success" is an instance where an individual uses social networking.

• Number of trials (n): 500
• Probability of success (p): 0.89 (as 89% of the sampled individuals use social networking)
• Number of successes (k): At least 95% of them, thus 475

This approach helps us understand the likelihood of 475 or more out of the 500 using social networking sites.

When dealing with large sample sizes, like 500 in our problem, computation using the binomial distribution formula can be complex and taxing. The normal approximation method provides a simpler alternative.

Normal approximation refers to using a normal distribution curve to estimate the probabilities of a binomial distribution, especially when the number of trials is large and both np and n(1-p) are greater than 5. These conditions hold true in our exercise:

• np = 500 * 0.89 = 445
• n(1-p) = 500 * 0.11 = 55

Both values are significantly greater than 5. Consequently, we use the normal approximation for easier calculations.

By approximating our discrete binomial dataset with a continuous normal distribution, we can use statistical tools involving the normal curve, enhancing our ability to compute probabilities smoothly.

The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1 and is typically denoted by the letter Z. It's a tool statisticians use to make comparisons across different data sets more straightforward.

In the context of this exercise, after determining the mean ( µ) and standard deviation ( σ) for our normal approximation, we convert our original problem to a standard normal problem. We aim to understand how far, in standard deviation units, our observed sample proportion (at least 95%) is from the mean proportion of 89%.

• Mean (µ): 445
• Standard deviation (σ): Approximately 7.38

Converting to the standard normal distribution through the process of standardization allows us to then consult standard normal distribution tables or use calculators for determining probabilities.

A Z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. Calculating the Z-score lets students standardize differences and understand probabilities concerning the standard normal distribution more effectively.

To find the Z-score in this exercise, we use the formula:\[ Z = \frac{X - \mu}{\sigma} \]Where:

• \( X \) is the observed value (475 in our case).
• \( \mu \) is the mean (445).
• \( \sigma \) is the standard deviation (7.38).

Once you calculate the Z-score, which approximately equals 4.07, you can use it to find the corresponding probability in the standard normal distribution table. In this situation, a Z-score of 4.07 suggests a probability that is exceptionally low, indicating it's rare for 95% or more young adults in a sample of 500 to use social networking, considering the baseline is 89%.