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assume the mean body fat percent for adult men is μ=18.94, the standard deviation is σ=7.75 and that the body fat distribution is normal.
) What is the probability that if a random sample of 90 men is chosen, the mean body fat percentage of the sample would be higher than 25%? assume the mean body fat percent for adult men is μ=18.94, the standard deviation is σ=7.75 and that the body fat distribution is normal.
) What is the probability that if a random sample of 90 men is chosen, the mean body fat percentage of the sample would be higher than 25%? @Statistics for th…
 2 years ago
 2 years ago
assume the mean body fat percent for adult men is μ=18.94, the standard deviation is σ=7.75 and that the body fat distribution is normal. ) What is the probability that if a random sample of 90 men is chosen, the mean body fat percentage of the sample would be higher than 25%? assume the mean body fat percent for adult men is μ=18.94, the standard deviation is σ=7.75 and that the body fat distribution is normal. ) What is the probability that if a random sample of 90 men is chosen, the mean body fat percentage of the sample would be higher than 25%? @Statistics for th…
 2 years ago
 2 years ago

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mainaknagBest ResponseYou've already chosen the best response.1
Let \[\theta\] be the mean of the sample of Size n. when, \[Z = (\theta  \mu) * \sqrt{n} / \sigma\] then, Z ~ N(0,1). (Please let me know if you need the proof) Now, for the sample mean to be > 0.25, we can use the standard normal distribution of Z, The probability can be given by P(Z > 0.07418), which is \[P(Z>0.07418) = 1  \phi (0.07418)\] Using \[\phi (0.07) = 0.52790 \] and \[\phi (0.08) = 0.53188\] and using the method of Interpolation  \[\phi (0.07418) = 0.5296\] Hence the Desired probability is 1  0.5296 = 0.4704
 2 years ago
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