SOLUTION: In university’s football team average weight of players is around (60 + α) kg, and the standard deviation is 4.37 lbs. Assuming heights are normally distributed (a) What perce

Question 1185548: In university’s football team average weight of players is around (60 + α) kg, and the standard
deviation is 4.37 lbs. Assuming heights are normally distributed
(a) What percentage of players are heavier than 70 kg?
(b) If your favorite player is within the heaviest (α+12)% of all players, what can his weight be?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
**(a) Percentage of players heavier than 70 kg:**
1. **Convert units:** The standard deviation is given in pounds, but the weight we're interested in is in kilograms. We need to convert either the standard deviation to kilograms or the weight to pounds. Since α is given in kg, and the mean is in kg, we'll convert the standard deviation to kg. 1 lb is approximately equal to 0.453592 kg. So, the standard deviation in kg is 4.37 lbs * 0.453592 kg/lb ≈ 1.98 kg.
2. **Calculate the z-score:** The z-score tells us how many standard deviations a value is away from the mean.
z = (x - μ) / σ
where:
* x is the value we're interested in (70 kg)
* μ is the mean weight (60 + α kg)
* σ is the standard deviation (1.98 kg)
z = (70 - (60 + α)) / 1.98 = (10 - α) / 1.98
3. **Find the probability:** Once we have the z-score, we can use a standard normal distribution table (a z-table) or calculator to find the probability of a player being heavier than 70 kg. We're looking for P(X > 70), where X is the weight of a randomly selected player. This is equivalent to finding the area to the *right* of the calculated z-score on the standard normal curve.
P(X > 70) = P(Z > z) = 1 - P(Z < z)
We find P(Z < z) from the z-table and subtract from 1.
**(b) Weight of the favorite player:**
1. **Find the z-score:** We know the player is within the heaviest (α + 12)% of players. This means that the area to the *right* of their weight on the normal distribution curve is (α + 12)/100. We need to find the z-score that corresponds to this area. We can use a z-table in reverse: look up the area in the *right tail* and find the corresponding z-score.
2. **Calculate the weight:** Once we have the z-score, we can use the z-score formula to solve for the player's weight (x):
z = (x - μ) / σ
x = μ + zσ
x = (60 + α) + z * 1.98
**Example (Let α = 4):**
**(a)**
1. σ ≈ 1.98 kg (as calculated above)
2. z = (10 - 4) / 1.98 ≈ 3.03
3. P(X > 70) = 1 - P(Z < 3.03). From a z-table, P(Z < 3.03) ≈ 0.9988. So, P(X > 70) ≈ 1 - 0.9988 = 0.0012, or about 0.12%.
**(b)**
1. Heaviest (4 + 12)% = 16% of players. We want the weight that separates the top 16% from the rest. In a z-table, we look for an area of 0.16 in the right tail. The corresponding z-score is approximately 1.0.
2. x = (60 + 4) + 1.0 * 1.98 ≈ 65.98 kg.
Therefore, if α = 4, about 0.12% of players are heavier than 70 kg, and a player in the top 16% weighs at least about 66 kg. You can substitute your actual value of α to get the final answers.

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