With the given information, we can use the properties of a normal distribution to estimate the percentage of widget weights that lie between 52 and 77 ounces.
Since the distribution is bell-shaped and we know the mean is 62 ounces and the standard deviation is 5 ounces, we can calculate the z-scores for the lower and upper bounds.
The z-score for 52 ounces is (52 - 62) / 5 = -2, and the z-score for 77 ounces is (77 - 62) / 5 = 3.
Using a standard normal distribution table or calculator, we can find the area under the curve between these z-scores.
The percentage of widget weights that lie between 52 and 77 ounces is approximately 99.73% or 99.7%.
Note:-
- Please note that this is an approximation based on the assumption of a normal distribution
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