Questions: For women aged 18-24, systolic blood pressures are normally distributed with a mean of 114.8 mm Hg and a standard deviation of 13.1 mm Hg. If 23 women aged 18-24 are randomly selected, find the probability that their mean systolic blood pressure is between 119 and 122 mm Hg.

Solution Steps

We are tasked with finding the probability that the mean systolic blood pressure of a sample of 23 women aged 18-24 falls between 119 mm Hg and 122 mm Hg. The systolic blood pressures are normally distributed with a mean (\( \mu \)) of 114.8 mm Hg and a standard deviation (\( \sigma \)) of 13.1 mm Hg.

To find the probability, we first convert the sample mean bounds (119 mm Hg and 122 mm Hg) into Z-scores using the formula:

\[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} \]

Where:

• \( X \) is the value for which we are calculating the Z-score,
• \( \mu = 114.8 \),
• \( \sigma = 13.1 \),
• \( n = 23 \).

Calculating the Z-scores:

• For the lower bound (119 mm Hg): \[ Z_{start} = \frac{119 - 114.8}{13.1 / \sqrt{23}} \approx 1.5376 \]

• For the upper bound (122 mm Hg): \[ Z_{end} = \frac{122 - 114.8}{13.1 / \sqrt{23}} \approx 2.6359 \]

The probability that the sample mean falls between the two Z-scores is given by:

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

Where \( \Phi \) is the cumulative distribution function (CDF) of the standard normal distribution. Using the calculated Z-scores:

\[ P = \Phi(2.6359) - \Phi(1.5376) \approx 0.0579 \]

Final Answer