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Answer :
We are given that the mean weight is [tex]$\mu = 172$[/tex] pounds and the standard deviation is [tex]$\sigma = 43$[/tex] pounds. We want to estimate the percentage of adult males that weigh more than 215 pounds.
Step 1. Calculate the [tex]$z$[/tex]-score
The [tex]$z$[/tex]-score is calculated using the formula
[tex]$$
z = \frac{x - \mu}{\sigma},
$$[/tex]
where [tex]$x$[/tex] is the weight threshold. Here, [tex]$x = 215$[/tex] pounds. Thus,
[tex]$$
z = \frac{215 - 172}{43} = \frac{43}{43} = 1.
$$[/tex]
Step 2. Apply the Empirical Rule
The Empirical Rule tells us that for a normal distribution:
- About 68% of the data falls within one standard deviation of the mean,
- About 95% falls within two standard deviations,
- About 99.7% falls within three standard deviations.
Since we are interested in weights more than one standard deviation above the mean (i.e., [tex]$z > 1$[/tex]), note that 68% of the data is within [tex]$\mu \pm \sigma$[/tex]. This means that
[tex]$$
100\% - 68\% = 32\%
$$[/tex]
of the data is outside this interval. Because the normal distribution is symmetric, half of these values lie above one standard deviation from the mean. Therefore, the percentage of data with [tex]$z > 1$[/tex] is approximately
[tex]$$
\frac{32\%}{2} = 16\%.
$$[/tex]
Conclusion
Thus, using the Empirical Rule, we estimate that approximately [tex]$16\%$[/tex] of adult males weigh more than 215 pounds.
The correct answer is [tex]$\boxed{16\%}$[/tex].
Step 1. Calculate the [tex]$z$[/tex]-score
The [tex]$z$[/tex]-score is calculated using the formula
[tex]$$
z = \frac{x - \mu}{\sigma},
$$[/tex]
where [tex]$x$[/tex] is the weight threshold. Here, [tex]$x = 215$[/tex] pounds. Thus,
[tex]$$
z = \frac{215 - 172}{43} = \frac{43}{43} = 1.
$$[/tex]
Step 2. Apply the Empirical Rule
The Empirical Rule tells us that for a normal distribution:
- About 68% of the data falls within one standard deviation of the mean,
- About 95% falls within two standard deviations,
- About 99.7% falls within three standard deviations.
Since we are interested in weights more than one standard deviation above the mean (i.e., [tex]$z > 1$[/tex]), note that 68% of the data is within [tex]$\mu \pm \sigma$[/tex]. This means that
[tex]$$
100\% - 68\% = 32\%
$$[/tex]
of the data is outside this interval. Because the normal distribution is symmetric, half of these values lie above one standard deviation from the mean. Therefore, the percentage of data with [tex]$z > 1$[/tex] is approximately
[tex]$$
\frac{32\%}{2} = 16\%.
$$[/tex]
Conclusion
Thus, using the Empirical Rule, we estimate that approximately [tex]$16\%$[/tex] of adult males weigh more than 215 pounds.
The correct answer is [tex]$\boxed{16\%}$[/tex].
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