If you replace the outlier of 113°F with 100°F, what happens to the range? Why do you think this occurs?
1 answer:
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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