Questions: Some sources report that the weights of full-term newborn babies in a certain town have a mean of 9 pounds and a standard deviation of 1.2 pounds and are Normally distributed. a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the mean-that is, between 7.8 and 10.2 pounds, or within one standard deviation of the mean? b. What is the probability that the average of four babies' weights will be within 1.2 pounds of the mean; will be between 7.8 and 10.2 pounds? c. Explain the difference between (a) and (b). a. The probability is 0.6827 . (Round to four decimal places as needed.) b. The probability is 0.9545 . (Round to four decimal places as needed.) c. The distribution of means is and than the original distribution. Therefore, the distribution of means will have observations located closer to the center of the distribution.

Some sources report that the weights of full-term newborn babies in a certain town have a mean of 9 pounds and a standard deviation of 1.2 pounds and are Normally distributed.
a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the mean-that is, between 7.8 and 10.2 pounds, or within one standard deviation of the mean?
b. What is the probability that the average of four babies' weights will be within 1.2 pounds of the mean; will be between 7.8 and 10.2 pounds?
c. Explain the difference between (a) and (b).
a. The probability is 0.6827 .
(Round to four decimal places as needed.)
b. The probability is 0.9545 .
(Round to four decimal places as needed.)
c. The distribution of means is and than the original distribution. Therefore, the distribution of means will have observations located closer to the center of the distribution.

Transcript text: Some sources report that the weights of full-term newborn babies in a certain town have a mean of 9 pounds and a standard deviation of 1.2 pounds and are Normally distributed. a. What is the probability that one newborn baby will have a weight within 1.2 pounds of the mean-that is, between 7.8 and 10.2 pounds, or within one standard deviation of the mean? b. What is the probability that the average of four babies' weights will be within 1.2 pounds of the mean; will be between 7.8 and 10.2 pounds? c. Explain the difference between (a) and (b). a. The probability is 0.6827 . (Round to four decimal places as needed.) b. The probability is 0.9545 . (Round to four decimal places as needed.) c. The distribution of means is $\square$ and $\square$ than the original distribution. Therefore, the distribution of means will have $\square$ observations located closer to the center of the distribution.

Solution

Solution Steps

Step 1: Probability for One Newborn

To find the probability that one newborn baby will have a weight within 1.2 pounds of the mean (between 7.8 and 10.2 pounds), we calculate the Z-scores for the bounds:

\[ Z_{start} = \frac{7.8 - 9.0}{1.2} = -1.0 \] \[ Z_{end} = \frac{10.2 - 9.0}{1.2} = 1.0 \]

Using the cumulative distribution function $ \Phi $, we find:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.0) - \Phi(-1.0) = 0.6827 \]

Thus, the probability that one newborn baby will weigh between 7.8 and 10.2 pounds is:

\[ \boxed{P = 0.6827} \]

Step 2: Probability for the Average of Four Newborns

Next, we calculate the probability that the average weight of four newborns will be within 1.2 pounds of the mean. The standard deviation of the sample mean is given by:

\[ \sigma_{mean} = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{4}} = 0.6 \]

We calculate the Z-scores for the same bounds:

\[ Z_{start} = \frac{7.8 - 9.0}{0.6} = -2.0 \] \[ Z_{end} = \frac{10.2 - 9.0}{0.6} = 2.0 \]

Using the cumulative distribution function $ \Phi $, we find:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(2.0) - \Phi(-2.0) = 0.9545 \]

Thus, the probability that the average weight of four newborns will be between 7.8 and 10.2 pounds is:

\[ \boxed{P = 0.9545} \]

Step 3: Explanation of Differences

The distribution of means is narrower and more concentrated than the original distribution. This is due to the Central Limit Theorem, which states that the distribution of the sample mean will have a smaller standard deviation (standard error) than the original distribution. Therefore, the distribution of means will have more observations located closer to the center of the distribution.