Question

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 a. 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). a. The percentage of bone density scores that are significantly high is nothing​%. ​(Round to two decimal places as​ needed.)

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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