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Answer :
To solve this problem, we need to find the probability that the combined weight of 36 men is more than 6,000 pounds. Given the mean and standard deviation of individual weights, we can use properties of the normal distribution and the Central Limit Theorem to approach this.
Identify the Mean and Standard Deviation of One Man's Weight:
- Mean weight, [tex]\mu = 163[/tex] pounds
- Standard deviation, [tex]\sigma = 18[/tex] pounds
Calculate the Mean and Standard Deviation for the Sample (36 Men):
According to the Central Limit Theorem, the sampling distribution of the sample mean will also be normally distributed for large [tex]n[/tex].- Mean of the sample, [tex]\mu_{sample} = \mu = 163[/tex] pounds
- Standard deviation of the sample (Standard Error), [tex]\sigma_{sample} = \frac{\sigma}{\sqrt{n}} = \frac{18}{\sqrt{36}} = 3[/tex] pounds
Find the Probability of the Combined Weight Being More Than 6,000 Pounds:
First, we find the mean of the total weight for 36 men:- Total mean weight, [tex]\mu_{total} = 36 \times \mu = 36 \times 163 = 5868[/tex] pounds
Then, find the standard deviation of the total weight:
- Total standard deviation, [tex]\sigma_{total} = 36 \times 3 = 108[/tex] pounds
Calculate the Z-score:
We need to calculate the [tex]z[/tex]-score for 6,000 pounds:[
z = \frac{X - \mu_{total}}{\sigma_{total}} = \frac{6000 - 5868}{108} = \frac{132}{108} \approx 1.22
]
Find the Probability:
Using standard normal distribution tables or a calculator, find the probability that [tex]Z > 1.22[/tex].- The probability corresponding to [tex]Z < 1.22[/tex] is approximately 0.8888.
- Therefore, the probability that [tex]Z > 1.22[/tex] is [tex]1 - 0.8888 = 0.1112[/tex].
So, the probability that the combined gross weight of 36 men traveling on a plane is more than 6,000 pounds is approximately 0.1112, or 11.12%.
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