Answer :
Final answer:
Using the principles of normal distribution and z-scores, we find that, from total 128 participants, approximately 91 people are estimated to have lost at least 15 pounds during the weight loss competition.
Explanation:
In this problem, we will utilize concepts of normal distribution and z-scores to identify the number of individuals who lost at least 15 pounds. The z-score represents the number of standard deviations an observation lies from the mean. It is calculated as z = (X – μ) / σ, where X is the observed value, μ is the mean, and σ is the standard deviation.
For a weight loss of 15 pounds, we calculate the corresponding z-score, z = (15 – 18) / 5.4 = -0.5556. We refer to a standard normal distribution table or calculator to find the proportion (p) of the distribution that lies below this z-score. This value is approximately 0.2891.
However, the question inquires about participants who lost 'at least 15 pounds', we need to quantify the proportion of the distribution that lies above this z-score. This is computed as 1 - p, which gives us 0.7109 or approximately 71.09%.
Finally, to find the number of people who lost at least 15 pounds, we multiply the total number of participants (128) by 0.7109. The resulting value is around 91, thus, it is estimated that about 91 participants lost at least 15 pounds.
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