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Answer :
To find the probability that the mean weight of 25 randomly selected players is between 180 lbs, we can use the properties of the normal distribution.
First, we identify the known values:
- Mean of the population, [tex]\mu = 182[/tex] lbs
- Standard deviation of the population, [tex]\sigma = 9.5[/tex] lbs
- Sample size, [tex]n = 25[/tex]
The mean of the sample distribution will also be [tex]\mu = 182[/tex] lbs. However, the standard deviation of the sample mean (known as the standard error) is calculated as:
[tex]\text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{9.5}{\sqrt{25}} = \frac{9.5}{5} = 1.9[/tex]
Now, we need to standardize the sample mean to use the standard normal distribution (Z-distribution). The formula for a Z-score is:
[tex]Z = \frac{\bar{X} - \mu}{\text{Standard Error}}[/tex]
We want to find the probability that the sample mean is less than 180 lbs:
- For 180 lbs:
[tex]Z_{180} = \frac{180 - 182}{1.9} = \frac{-2}{1.9} \approx -1.053[/tex]
Next, we find the probability associated with [tex]Z = -1.053[/tex]. Using the standard normal distribution table or a calculator, we find:
[tex]P(Z < -1.053) \approx 0.1469[/tex]
Therefore, the probability that the mean weight of the selected players is less than 180 lbs is approximately 0.1469, or 14.69%.
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