To solve this problem, we need to calculate the probability that a newborn baby’s weight falls within 1 standard deviation of the mean in a normally distributed population.

We know:

• The mean $\mu = 9$ pounds,
• The standard deviation $\sigma = 1.2$ pounds,
• The weight range between 7.8 pounds and 10.2 pounds is equivalent to being within 1 standard deviation of the mean (since $9 - 1.2 = 7.8$ and $9 + 1.2 = 10.2$).

Step 1: Standardize the values

To find the probability, we need to convert the given range into z-scores using the formula:

$z = \frac{x - \mu}{\sigma}$

For the lower bound ($x = 7.8$): $z_{\text{lower}} = \frac{7.8 - 9}{1.2} = \frac{-1.2}{1.2} = -1$

For the upper bound ($x = 10.2$): $z_{\text{upper}} = \frac{10.2 - 9}{1.2} = \frac{1.2}{1.2} = 1$

Step 2: Find the probability corresponding to these z-scores

For a standard normal distribution, the probability of a z-score falling between $-1$ and $1$ is approximately $0.6826$. This is a well-known result for the normal distribution.

Conclusion:

The probability that a newborn baby’s weight is between 7.8 pounds and 10.2 pounds (i.e., within 1 standard deviation of the mean) is 0.6826 or 68.26%.

Would you like further details on this process or additional questions to explore?

Here are five related questions for practice:

1. What is the probability of a newborn weighing more than 10.2 pounds?
2. What is the probability of a newborn weighing less than 7.8 pounds?
3. How would the probability change if the standard deviation were smaller?
4. What percentage of newborns would weigh between 6.6 pounds and 11.4 pounds (within 2 standard deviations of the mean)?
5. How would you calculate the probability for a newborn weighing more than 12 pounds?

Tip: When working with normal distributions, always standardize values to z-scores to use standard normal probability tables effectively.