Question

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer b. If the length of pregnancy is in the lowest 4%, then the baby is premature. Find the length that separates premature babies from those who are not premature Click to view Page 1 of the table. Click to view.ge 2 of the table. a. The probability that a pregnancy wil last 309 days or longer is I (Round to four decimal places as needed.) 390R19 Points: 0 of 2 Save Assume that females have pulse rates that are normally distributed with a mean of p = 72.0 beats per minute and a standard deviation of a = 125 beats per minute Complete parts (a) through (c) below. a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 65 beats per minute and 79 beats per minute. The probability is (Round to four decimal places as needed.)

Answer 1

The probability of a pregnancy lasting 309 days or longer is approximately 0.0031. The length that separates premature babies from those who are not premature is approximately 240.25 days.

Step-by-step explanation:

a. To find the probability of a pregnancy lasting 309 days or longer, we need to find the area under the normal distribution curve to the right of 309. First, we need to calculate the z-score using the formula z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. In this case, x = 309, μ = 268, and σ = 15. Plugging in these values, we get z = (309 - 268) / 15 ≈ 2.73. Now we can use a standard normal distribution table or a calculator to find the probability associated with a z-score of 2.73. Looking up the z-score in the table, we find that the probability is approximately 0.0031.

b. To find the length that separates premature babies from those who are not premature, we need to find the z-score corresponding to the lowest 4%. We can use a standard normal distribution table or a calculator to find the z-score associated with a cumulative probability of 0.04. Looking up the cumulative probability in the table, we find that the z-score is approximately -1.75. Now we can use the formula z = (x - μ) / σ to solve for x. Rearranging the formula, we get x = μ + z * σ. Plugging in the values, we get x = 268 + (-1.75) * 15 ≈ 240.25. So the length that separates premature babies from those who are not premature is approximately 240.25 days.