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Answer:
a) The percentage of bone density scores that are significantly high is 2.28%
b) The percentage of bone density scores that are significantly low is 2.28%
c) The percentage of bone density scores that are not significant is 95.44%
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
a. significantly high (or at least 2 standard deviations above the mean).
This is 1 subtracted by the pvalue of Z = 2.
Z = 2 has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% of scores are signifcantly high
b. significantly low (or at least 2 standard deviations below the mean).
pvalue of Z = -2
Z = -2 has a pvalue of 0.0228
2.28% of scores are signicantly low.
c. not significant (or less than 2 standard deviations away from the mean).
pvalue of Z = 2 subtracted by the pvalue of Z = -2.
Z = 2 has a pvalue of 0.9772
Z = -2 has a pvalue of 0.0228
0.9772 - 0.0228 = 0.9544
95.44% of the scores are not significant
we need to divide the population of Orlando Florida between the population of Jasper
the population of Orlando Florida is 3000 times larger than the population of Ja