Part 1 of 3 Points: 0 of 1 Save The height of women ages 20-29 is normally distributed, with a mean of 63.9 inches. Assume \( \sigma=2.7 \) inches. Are you more likely to randomly select 1 woman with a height less than 65.4 inches or are you more likely to select a sample of 18 women with a mean height less than 65.4 inches? Explain. Click the icon to view page 1 of the standard normal table. Click the icon to view page 2 of the standard normal table. What is the probability of randomly selecting 1 woman with a height less than 65.4 inches? \( \square \) (Round to four decimal places as needed.)

We are given that women's heights are normally distributed with a mean (μ) of 63.9 inches and a standard deviation (σ) of 2.7 inches. Step 1. Find the Z-score for a height of 65.4 inches using the formula:   Z = (X – μ) / σ    = (65.4 – 63.9) / 2.7    = 1.5 / 2.7    ≈ 0.5556 Step 2. Look up (or calculate) the probability P(Z < 0.5556) using the standard normal table or a calculator. This probability is approximately 0.7123. Thus, the probability of randomly selecting one woman with a height less than 65.4 inches is about 0.7123 (when rounded to four decimal places). For context, although not asked in this part, note that if we take a sample of 18 women, the standard error of the mean is σ/√n = 2.7/√18 ≈ 0.636. Then the Z-score for the sample mean being less than 65.4 inches is much higher (about 2.36), which results in a probability that is even greater. This means that it is more likely that the average height of a sample of 18 women is less than 65.4 inches than the height of one randomly selected woman is less than 65.4 inches. The answer to this part is: 0.7123.