Question

Assume that the average weight of an NFL player is 245.7 pounds
with a standard deviation of 34.5 pounds, but the probability
distribution is unknown.

a. if a random sample of 32 NFL players is selected, what is
the probabilty that the average weight of the sample sill be less
than 234 pounds?

b. If a random sample of 32 NFL players is selected, what is
the probability that the average weight of the sample is between
248 and 254 pounds?

c. what is the value of the standard error?

Answer #1

μ = 245.7, σ = 34.5

Even though the distribution is unknown, since n > 30, the sampling distribution of the mean is approximately normal

(a) z = (x-bar - μ)/(s/√n) = (234 - 245.7)/(34.5/√32) = -1.9184

P(x-bar < 234) = P(z < -1.9184) = 0.0275

(b) z1 = (248 - 245.7)/(34.5/√32) = -0.3771 and z2 = (254 - 245.7)/(34.5/√32) = 1.3609

P(248 < x-bar < 254) = P(-0.3771 < z < 1.3609) = 0.2663

(c) Standard error = s/√n = 34.5/√32 = 6.0988

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