Questions: For bone density scores that are normally distributed with a mean, 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 (Round to two decimal places as needed.)

Solution Steps

The mean \( \mu \) of the bone density scores is calculated as follows:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{12.6}{10} = 1.26 \]

Thus, the mean of bone density scores is:

Mean of bone density scores: \( \mu = 1.26 \)

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n} = 0.21 \]

The standard deviation \( \sigma \) is then:

\[ \sigma = \sqrt{0.21} \approx 0.45 \]

Thus, the standard deviation of bone density scores is:

Standard deviation of bone density scores: \( \sigma \approx 0.45 \)

To find the probability of scores being significantly high (at least 2 standard deviations above the mean), we calculate the Z-scores:

\[ Z_{start} = \frac{range\_start - \mu}{\sigma} = \frac{\mu + 2\sigma - \mu}{\sigma} = 2.0 \]

The probability \( P \) is given by:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(\infty) - \Phi(2.0) = 0.0228 \]

Thus, the percentage of bone density scores that are significantly high is:

\[ \text{Percentage} = (1 - P) \times 100 = (1 - 0.0228) \times 100 \approx 97.72\% \]

Final Answer