Questions: The mean weight gain for women during a full-term pregnancy is 30.2 pounds. The standard deviation of weight gain for this group is 9.9 pounds, and the shape of the distribution of weight gains is symmetric and unimodal. Complete parts (a) and (b) below.

Solution Steps

To solve the given problem, we need to understand the properties of the normal distribution since the distribution of weight gains is symmetric and unimodal. We can use the mean and standard deviation to calculate probabilities or percentiles related to the weight gain during pregnancy.

(a) To find the probability of a certain weight gain range, we can use the cumulative distribution function (CDF) of the normal distribution.

(b) To find a specific percentile of the weight gain distribution, we can use the inverse of the CDF, also known as the percent-point function (PPF).

We are given that the mean weight gain during a full-term pregnancy is \( \mu = 30.2 \) pounds, with a standard deviation \( \sigma = 9.9 \) pounds. The distribution is symmetric and unimodal, indicating a normal distribution.

To find the probability that the weight gain is less than 40 pounds, we use the cumulative distribution function (CDF) for a normal distribution. The probability is given by: \[ P(X < 40) = \Phi\left(\frac{40 - \mu}{\sigma}\right) \] Substituting the given values: \[ P(X < 40) = \Phi\left(\frac{40 - 30.2}{9.9}\right) \approx 0.8389 \]

The 90th percentile of a normal distribution is found using the inverse of the CDF, also known as the percent-point function (PPF). The 90th percentile \( x_{0.90} \) is given by: \[ x_{0.90} = \mu + z_{0.90} \cdot \sigma \] where \( z_{0.90} \) is the z-score corresponding to the 90th percentile. Substituting the given values: \[ x_{0.90} = 30.2 + 1.2816 \cdot 9.9 \approx 42.8874 \]

Final Answer