Questions: Suppose that the probability distribution for birth weights is normal with a mean of 120 ounces and a standard deviation of 20 ounces. The probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces is [ Select ] 68 %. The probability that a randomly selected infant has a birth weight between 110 and 130 is [Select] 68 %.

Solution Steps

To find the probability that a randomly selected infant has a birth weight between 100 ounces and 140 ounces, we calculate the Z-scores for the bounds:

\[ Z_{start} = \frac{100 - 120}{20} = -1.0 \] \[ Z_{end} = \frac{140 - 120}{20} = 1.0 \]

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 is:

\[ P \approx 68.27\% \]

Next, we calculate the probability that a randomly selected infant has a birth weight between 110 ounces and 130 ounces. The Z-scores for these bounds are:

\[ Z_{start} = \frac{110 - 120}{20} = -0.5 \] \[ Z_{end} = \frac{130 - 120}{20} = 0.5 \]

Again, using the cumulative distribution function \( \Phi \):

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.5) - \Phi(-0.5) = 0.3829 \]

Thus, the probability is:

\[ P \approx 38.29\% \]

Final Answer