The weight of male babies less than 2 months old in the United States is normally distributed with mean 12.5 pounds and standard deviation 5.1 pounds.

Question

A) what proportion of babies weigh more than 14 pounds?

Answer 1

To find the proportion of male babies less than 2 months old that weigh more than 14 pounds, we can use the properties of the normal distribution.

Given:

• Mean (\( \mu \)) = 12.5 pounds
• Standard deviation (\( \sigma \)) = 5.1 pounds

We can standardize the weight of 14 pounds to find the corresponding \( z \)-score using the formula:

\[ z = \frac{(X - \mu)}{\sigma} \]

where \( X \) is the value we are interested in (14 pounds).

Substituting the values:

\[ z = \frac{(14 - 12.5)}{5.1} = \frac{1.5}{5.1} \approx 0.2941 \]

Next, we can use the \( z \)-score to find the proportion of babies that weigh more than 14 pounds. We look up the \( z \)-score in the standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF).

A \( z \)-score of approximately 0.2941 typically corresponds to a cumulative probability of about 0.6151 (this value may vary slightly depending on the table or calculator used).

To find the proportion of babies that weigh more than 14 pounds, we subtract the cumulative probability from 1:

\[ P(X > 14) = 1 - P(Z < 0.2941) \approx 1 - 0.6151 = 0.3849 \]

Thus, the proportion of male babies less than 2 months old that weigh more than 14 pounds is approximately 0.3849, or 38.49%.

Answer 2

B) what proportion of babies weigh less than 16 pounds?

Answer 3

To find the proportion of male babies less than 2 months old that weigh less than 16 pounds, we will again use the properties of the normal distribution.

Given:

• Mean (\( \mu \)) = 12.5 pounds
• Standard deviation (\( \sigma \)) = 5.1 pounds

We can standardize the weight of 16 pounds to find the corresponding \( z \)-score using the formula:

\[ z = \frac{(X - \mu)}{\sigma} \]

where \( X \) is the value we are interested in (16 pounds).

Substituting the values:

\[ z = \frac{(16 - 12.5)}{5.1} = \frac{3.5}{5.1} \approx 0.6863 \]

Next, we can use the \( z \)-score to find the cumulative probability of weights less than 16 pounds. We look up the \( z \)-score in the standard normal distribution table or use a calculator that provides the cumulative distribution function (CDF).

A \( z \)-score of approximately 0.6863 typically corresponds to a cumulative probability of about 0.7549.

Therefore, the proportion of male babies less than 2 months old that weigh less than 16 pounds is approximately 0.7549, or 75.49%.

Answer 4

c) what proportion of babies weigh between 11 and 15 pounds?

Answer 5

To find the proportion of male babies less than 2 months old that weigh between 11 and 15 pounds, we will calculate the \( z \)-scores for both weights (11 pounds and 15 pounds) and then find the corresponding cumulative probabilities.

1. Calculate the \( z \)-score for 11 pounds:

\[ z_{11} = \frac{(X - \mu)}{\sigma} = \frac{(11 - 12.5)}{5.1} = \frac{-1.5}{5.1} \approx -0.2941 \]

2. Calculate the \( z \)-score for 15 pounds:

\[ z_{15} = \frac{(X - \mu)}{\sigma} = \frac{(15 - 12.5)}{5.1} = \frac{2.5}{5.1} \approx 0.4902 \]

3. Find the cumulative probabilities:

Next, we look up the cumulative probabilities for these \( z \)-scores:

• For \( z_{11} \approx -0.2941 \), the cumulative probability \( P(Z < -0.2941) \) is approximately 0.3859.
• For \( z_{15} \approx 0.4902 \), the cumulative probability \( P(Z < 0.4902) \) is approximately 0.6879.

4. Calculate the proportion between 11 and 15 pounds:

To find the proportion of babies that weigh between 11 and 15 pounds, we subtract the cumulative probability at 11 pounds from the cumulative probability at 15 pounds:

\[ P(11 < X < 15) = P(Z < 0.4902) - P(Z < -0.2941) \approx 0.6879 - 0.3859 = 0.3020 \]

Thus, the proportion of male babies less than 2 months old that weigh between 11 and 15 pounds is approximately 0.3020, or 30.20%.

Answer 6

is it unusual for a baby to weigh more than 17 pounds?
[yes or no] because the probability is [?]

Answer 7

To determine whether it is unusual for a baby to weigh more than 17 pounds, we first need to calculate the \( z \)-score for 17 pounds and then find the corresponding probability.

1. Calculate the \( z \)-score for 17 pounds:

Using the formula for the \( z \)-score:

\[ z = \frac{(X - \mu)}{\sigma} \]

For \( X = 17 \):

\[ z = \frac{(17 - 12.5)}{5.1} = \frac{4.5}{5.1} \approx 0.8824 \]

2. Find the cumulative probability:

Next, we look up the cumulative probability for \( z \approx 0.8824 \). This typically corresponds to a cumulative probability of about 0.8133.

3. Find the probability of a baby weighing more than 17 pounds:

To find the probability of a baby weighing more than 17 pounds, we subtract the cumulative probability from 1:

\[ P(X > 17) = 1 - P(Z < 0.8824) \approx 1 - 0.8133 = 0.1867 \]

Conclusion:

Since the probability of a baby weighing more than 17 pounds is approximately 0.1867 (or 18.67%), whether this is considered unusual can depend on a common statistical threshold. A value greater than 5% is often not deemed unusual.

Therefore, the answer is:

No, it is not unusual for a baby to weigh more than 17 pounds because the probability is approximately 0.1867 (or 18.67%).