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For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are
For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that area. significantly high (or at least 2 standard deviations above the mean).
b. significantly low (or at least 2 standard deviations below the mean).
c. not significant (or less than 2 standard deviations away from the mean).
Using thenormal distribution, it is found that:a) 2.28%of scores are significantly high.b) 2.28%of scores are significantly low.c) 95.44%of scores are not significant.In anormal distributionwithmeanandstandard deviation, thez-scoreof ameasure Xis given by:Itmeasureshow many standard deviations the measure is from the mean.After finding the z-score, we look at the z-score table and find thep-valueassociated with this z-score, which is thepercentileof both X and Z.Itema:Theprobabilityof finding a value at least 2 standard deviations above the mean is:, which is1 subtracted by the p-value of Z = 2.Z = 2 has a p-value of 0.9772.1 - 0.9772 = 0.0228Then,2.28%of scores are significantly high.Itemb:Theprobabilityof finding a value at least 2 standard deviations below the mean is:, which isthe p-value of Z = 2.Z = -2 has a p-value of 0.0228.Then,2.28%of scores are significantly low.Itemc:Theprobabilityis, which is thep-value of Z = 2 subtracted by the p-value of Z = -2.Z = 2 has a p-value of 0.9772.Z = -2 has a p-value of 0.0228.Then,95.44%of scores are not significant.A similar problem is given atbrainly.com/question/13073931...