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The Z-score is a measure that indicates how many standard deviations a data point is from the mean of a data set. In the context of a normal distribution, it allows you to determine the probability of a certain data point within that distribution. When calculating a Z-score, you use the formula:

• \( Z = \frac{X - \mu}{\sigma} \)

Here, \( X \) represents the data point of interest, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation.
To find the probability of at least 30% of 18 to 34-year-olds checking Facebook in the morning, we first calculate \( Z \):

• Substitute: \( X = 30 \), \( \mu = 28 \), \( \sigma = 5 \)
• Calculate \( Z = \frac{30 - 28}{5} = 0.4 \)

By finding this Z-score, we establish where this falls in terms of standard deviation from the mean. It helps in looking up probabilities in the Z-table and understanding the data location within the distribution.

Probability analysis involves understanding the likelihood of various outcomes within a probability distribution. In this exercise, the normal distribution is key, as it often appears in natural phenomena and everyday situations due to the Central Limit Theorem.
Once you have the Z-score, you can identify the probability of a data point being less than a certain value using a Z-table. With a Z-score of 0.4, the table shows us:

• \( P(Z < 0.4) \approx 0.6554 \)

This value suggests there's a 65.54% chance that the percentage of people who check Facebook is less than 30%. To find the counterpart probability of at least 30%, subtract from 1:

• \( P(Z > 0.4) = 1 - 0.6554 = 0.3446 \)

Thus, there's a 34.46% probability that the percentage is at least 30%.

Percentiles express the relative standing of a value within a data set. The 95th percentile, for instance, means that 95% of values fall below this point. It is particularly useful in determining thresholds or comparing where a particular observation stands within a distribution.
For this exercise, we want to find the percentage reflecting the 95th percentile of individuals checking Facebook before getting out of bed. Using a Z-table or calculator, we find:

• The Z-score for the 95th percentile is approximately 1.65.

Then, solving for \( X \) using the Z-score formula in reverse:

• \( X = Z \times \sigma + \mu \)
• \( X = 1.65 \times 5 + 28 = 36.25 \)

So, 95% of the younger population checks Facebook at a rate less than 36.25%. This means that if you randomly select an individual from this group, there's a 95% chance they check Facebook less often than this threshold.