QuestionJuly 15, 2025

deviatio that the cholesterol levels of adult American women can be described by a normal model with a mean of 188mg/dL and a star deviation of 24mg/dL Based on this assumption what percent of adult American women do you expect to have cholesterol levels between 150mg/dL and mg/dL? (express your answer to two decimal places) Percentage=type your answer... %

deviatio that the cholesterol levels of adult American women can be described by a normal model with a mean of 188mg/dL and a star deviation of 24mg/dL Based on this assumption what percent of adult American women do you expect to have cholesterol levels between 150mg/dL and mg/dL? (express your answer to two decimal places) Percentage=type your answer... %
deviatio that the cholesterol levels of adult American women can be described by a normal model with a mean of 188mg/dL and a star deviation of 24mg/dL Based on this assumption what percent of adult American women do you expect to have cholesterol levels between 150mg/dL and mg/dL? (express your answer to two decimal places) Percentage=type your answer... %

Solution
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Answer

63.44\% Explanation 1. Calculate Z-scores Use the formula for Z-score: Z = \frac{X - \mu}{\sigma}. For X = 150, Z_1 = \frac{150 - 188}{24} = -1.58. For X = 200, Z_2 = \frac{200 - 188}{24} = 0.50. 2. Find probabilities from Z-table Look up Z_1 = -1.58 and Z_2 = 0.50 in the standard normal distribution table. P(Z < -1.58) \approx 0.0571 and P(Z < 0.50) \approx 0.6915. 3. Calculate percentage between Z-scores Subtract the probabilities: P(-1.58 < Z < 0.50) = P(Z < 0.50) - P(Z < -1.58) = 0.6915 - 0.0571 = 0.6344. 4. Convert to percentage Multiply by 100 to convert to percentage: 0.6344 \times 100 = 63.44\%.

Explanation

1. Calculate Z-scores<br /> Use the formula for Z-score: $Z = \frac{X - \mu}{\sigma}$. For $X = 150$, $Z_1 = \frac{150 - 188}{24} = -1.58$. For $X = 200$, $Z_2 = \frac{200 - 188}{24} = 0.50$.<br /><br />2. Find probabilities from Z-table<br /> Look up $Z_1 = -1.58$ and $Z_2 = 0.50$ in the standard normal distribution table. $P(Z < -1.58) \approx 0.0571$ and $P(Z < 0.50) \approx 0.6915$.<br /><br />3. Calculate percentage between Z-scores<br /> Subtract the probabilities: $P(-1.58 < Z < 0.50) = P(Z < 0.50) - P(Z < -1.58) = 0.6915 - 0.0571 = 0.6344$.<br /><br />4. Convert to percentage<br /> Multiply by 100 to convert to percentage: $0.6344 \times 100 = 63.44\%$.
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