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The weight of male babies less than 2 months old in the United States is normally distributed with mean 12.3 and standard deviation 4.7. Round the answers to four decimal places.

(a) What proportion of babies weigh more than 12 pounds ?

7 answers

To find the proportion of male babies less than 2 months old who weigh more than 12 pounds, we will first convert the weight into a z-score using the given mean and standard deviation.

The formula for the z-score is:

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

where:

  • \(X\) is the value (12 pounds in this case),
  • \(\mu\) is the mean (12.3 pounds),
  • \(\sigma\) is the standard deviation (4.7 pounds).

Substituting the values:

\[ z = \frac{12 - 12.3}{4.7} = \frac{-0.3}{4.7} \approx -0.0638 \]

Next, we will look up the z-score in the standard normal distribution table or use a calculator to find the proportion of values less than this z-score.

Using the standard normal distribution table, or a calculator:

  • The cumulative probability (area to the left of \(z = -0.0638\)) is approximately \(0.4756\).

To find the proportion of babies weighing more than 12 pounds, we calculate:

\[ P(X > 12) = 1 - P(X \leq 12) \]

Thus,

\[ P(X > 12) = 1 - 0.4756 \approx 0.5244 \]

So the proportion of babies who weigh more than 12 pounds is approximately:

\[ \boxed{0.5244} \]

(b) What proportion of babies weigh less than 15 pounds ?

To find the proportion of male babies less than 2 months old who weigh less than 15 pounds, we'll again use the z-score formula:

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

where:

  • \(X\) is the value (15 pounds in this case),
  • \(\mu\) is the mean (12.3 pounds),
  • \(\sigma\) is the standard deviation (4.7 pounds).

Substituting the values:

\[ z = \frac{15 - 12.3}{4.7} = \frac{2.7}{4.7} \approx 0.5745 \]

Next, we will look up the z-score in the standard normal distribution table or use a calculator to find the cumulative probability.

Using a standard normal distribution table or a calculator, the cumulative probability for \(z = 0.5745\) is approximately \(0.7181\).

Therefore, the proportion of babies who weigh less than 15 pounds is approximately:

\[ \boxed{0.7181} \]

(c) What proportion of babies weigh between 9 and 13 pounds ?

To find the proportion of male babies less than 2 months old who weigh between 9 and 13 pounds, we will calculate the z-scores for both weights and then find the cumulative probabilities.

Step 1: Calculate the z-scores

  1. For 9 pounds:

\[ z = \frac{X - \mu}{\sigma} = \frac{9 - 12.3}{4.7} = \frac{-3.3}{4.7} \approx -0.7021 \]

  1. For 13 pounds:

\[ z = \frac{X - \mu}{\sigma} = \frac{13 - 12.3}{4.7} = \frac{0.7}{4.7} \approx 0.1489 \]

Step 2: Find cumulative probabilities

Using the standard normal distribution table or a calculator:

  1. Cumulative probability for \(z = -0.7021\) is approximately \(0.2420\).
  2. Cumulative probability for \(z = 0.1489\) is approximately \(0.5596\).

Step 3: Calculate the proportion between the two weights

To find the proportion of babies weighing between 9 and 13 pounds, we subtract the cumulative probability of \(z = -0.7021\) from that of \(z = 0.1489\):

\[ P(9 < X < 13) = P(X < 13) - P(X < 9) = 0.5596 - 0.2420 = 0.3176 \]

Thus, the proportion of babies who weigh between 9 and 13 pounds is approximately:

\[ \boxed{0.3176} \]

d) Is it unusual for a baby to weigh more than 18.1 pounds ? why or why not

To determine if it is unusual for a baby to weigh more than 18.1 pounds, we can calculate the z-score for 18.1 pounds and then analyze the corresponding cumulative probability.

Step 1: Calculate the z-score

Using the formula for the z-score:

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

where:

  • \(X = 18.1\) pounds,
  • \(\mu = 12.3\) pounds (mean),
  • \(\sigma = 4.7\) pounds (standard deviation).

Substituting the values:

\[ z = \frac{18.1 - 12.3}{4.7} = \frac{5.8}{4.7} \approx 1.2340 \]

Step 2: Find the cumulative probability

Using a z-table or calculator, we can find the cumulative probability for \(z = 1.2340\):

  • The cumulative probability for \(z = 1.2340\) is approximately \(0.8907\).

This means that approximately 89.07% of babies weigh less than 18.1 pounds.

Step 3: Determine if it is unusual

To determine if a weight of more than 18.1 pounds is unusual, we can consider the following guideline:

  • If a value falls outside of 2 standard deviations from the mean (which encompasses about 95% of the data in a normal distribution), it may be considered unusual.

Calculating the cutoff for more than 18.1 pounds:

\[ P(X > 18.1) = 1 - P(X < 18.1) = 1 - 0.8907 = 0.1093 \]

This means that about 10.93% of the population weighs more than 18.1 pounds.

Conclusion

Since the proportion of babies weighing more than 18.1 pounds (approximately 10.93%) is greater than 5% but significantly less than 50%, it is not extremely unusual, but it may still be considered somewhat unusual since it lies beyond the 90th percentile threshold (which indicates that only 10% of babies weigh more than this amount).

Therefore, while it is not extremely unusual for a baby to weigh more than 18.1 pounds, it is relatively uncommon when considering the general distribution of weights.