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assume that a randomly selected subject is given a bone density test. those test scores are normally distributed with a mean of 0 and a standard deviation of 1. draw a graph and find the probability of a bone density test score greater than .

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The probability that a bone density test score greater than 0.49 calculated using the Zscore formula is 0.312

Given the Parameters :

  • Mean, μ = 0
  • Standard deviation, σ = 1
  • Score, X = 0.49

Recall, the Z score formula :

  • Zscore = (X - mean) / standard deviation

Zscore = (0.49 - 0) / 1 = 0.49

The Probability can be expressed thus :

P(Z > 0.49) = 1 - P(Z < 0.49)

Using a probability calculator or normal distribution table ;

P(Z < 0.49) = 0.68793

P(Z > 0.49) = 1 - 0.68793

P(Z > 0.49) = 0.31207

Therefore, the P(Z > 0.49) will be 0.312

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