Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

elizbaker45

  • 4 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…

  • This Question is Closed
  1. mainaknag
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy