You are interested in studying social media use among the elderly. Through previous research, you know that the sampling distribution of the mean of your social media use measure has a mean of 3.2 (hours per week) and a standard deviation of 1.6. You plan to take a large random sample of elderly individuals. What is the probability that the mean level of social media use in your sample will be 4 hours per week or more?

To find the probability that the mean level of social media use in the sample will be 4 hours per week or more, we need to calculate the Z-score for 4 hours and then find the corresponding probability from the standard normal distribution.

### Step 1: Calculate the Z-score
The formula for the Z-score is:

\[
Z = \frac{X - \mu}{\sigma / \sqrt{n}}
\]

Where:
- \(X\) is the sample mean (4 hours)
- \(\mu\) is the population mean (3.2 hours)
- \(\sigma\) is the population standard deviation (1.6 hours)
- \(n\) is the sample size

Since a large random sample is being taken, we can use the Central Limit Theorem, which states that the sampling distribution of the sample mean is approximately normal for large \(n\). We can proceed without knowing the exact sample size because the standard error will still be valid as long as \(n\) is large.

### Step 2: Plug in the values
Let's rewrite the Z-score formula:

\[
Z = \frac{4 - 3.2}{1.6 / \sqrt{n}}
\]

Since the sample size \(n\) is large, we can assume that the standard error of the mean (\(\sigma / \sqrt{n}\)) approaches zero. Therefore, we can look up the Z value using the standard normal distribution tables or software for exact probabilities.

Explore the probability that the mean level of social media use in a large sample of elderly individuals will be 4 hours or more. The problem involves key statistical concepts such as the Z-score formula and the Central Limit Theorem. Learn how to calculate probabilities using the standard normal distribution with a population mean of 3.2 hours and standard deviation of 1.6 hours.

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