Assume that round widgets from a manufacturing process have diameters that are normally distributed with mean 175.25 centimeters and standard deviation centimeters. The chances that a randomly selected widget has diameter less than 166 centimeters is closest to which of the following? a. 1 out of 5000 b. 1 out of 50000 c. 1 out of 1000 d. 1 out of 25000

To find the probability that a randomly selected widget has a diameter less than 166 centimeters, we first need to standardize the value using the Z-score formula: \( Z = \frac{(X - \mu)}{\sigma} \), where \( X \) is the diameter (166 cm), \( \mu \) is the mean (175.25 cm), and \( \sigma \) is the standard deviation. Assuming the standard deviation is, say, 10 cm for demonstration, the calculation would be \( Z = \frac{(166 - 175.25)}{10} \approx -0.925 \). Using Z-tables or calculators, we can find that the probability of a Z-score of -0.925 corresponds to approximately 0.1788. Hence, \( P(X < 166) \approx 0.1788\), which means that the chances of selecting a widget with a diameter less than 166 cm are about 1 out of 5.6, not matching any of the options provided. An updated estimate may yield different results with an unknown standard deviation, so you may want to adjust that figure accordingly. If we think about real-life production, knowing the diameter of widgets can help manufacturers optimize their quality control processes. If too many widgets fall outside of specifications (either too small or too large), it can lead to increased costs and customer dissatisfaction. Thus, using statistical analysis not only helps in forecasting quality but also in making informed decisions regarding production adjustments, ensuring that the final products meet consumer expectations while minimizing waste.