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Answer :
Option B, The center 95% of players had an average weight of 151 pounds.
Football players' weights have a mean of 200 pounds as well as a standard deviation of 25 pounds, which corresponds to a normal distribution.
So, let x represent a football player's weight.
X ≅ N(200, (25)²)
The minimal weight of the center 95% of the members is
P(X > 95) = P((95 - 200) ÷ 25) < (X - µ) ÷ σ)
P(X > 95) = P(-4.2 < z)
P(X > 95) = - P(z < -4.2)
P(X > 95) = 0.1512
The needed percentage is 0.1512 × 100 = 151.2 percent.
A probability distribution that is equal around the mean is the normal distribution, sometimes referred to as the Gaussian distribution. It demonstrates that data that are close to the mean occur more frequently than data that are far from the mean.
Learn more about normal distribution at
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