4.6.2 Test (CST): Statistical Analysis Question 12 of 25 Suppose a normal distribution has a mean of 79 and a standard deviation of 7. What is \( P(x \geq 72) \) ? \( \begin{array}{l}\text { A. } 0.16 \\ \text { B. } 0.025 \\ \text { C. } 0.975 \\ \text { D. } 0.84 \\ \text { 1 SuBMIT }\end{array} \)

To find \( P(x \geq 72) \) for a normally distributed variable with a mean (\( \mu \)) of 79 and a standard deviation (\( \sigma \)) of 7, we first convert the value 72 into a z-score using the formula \( z = \frac{(x - \mu)}{\sigma} \). So, \( z = \frac{(72 - 79)}{7} = \frac{-7}{7} = -1 \). Now, we look up the z-score of -1 in the standard normal distribution table, which gives us \( P(Z < -1) \approx 0.1587 \). Therefore, \( P(Z \geq -1) = 1 - P(Z < -1) \approx 0.8413 \). Rounding it a bit, we find that \( P(x \geq 72) \) is approximately 0.84, so the answer is D. Here's some fun extra info: When it comes to statistics, the normal distribution is like the superstar of probability models! It’s used in everything from determining test scores to analyzing heights and weights. Its symmetric bell shape helps us visualize how data is spread around the mean, making it easier for statisticians to analyze patterns in various real-world situations. Ever find yourself confused with z-scores? Don’t worry; it’s a common pitfall! A frequent mistake is forgetting to subtract the mean from your x-value before dividing by the standard deviation. Also, don’t skip the step of looking up z-scores in the table! Each of these steps is crucial to ensure you accurately interpret the results and avoid any embarrassing errors on your test.