In a particular country, the height measurements of ten-year-old children are approximately normally distributed with a mean of 54.8 inches, and standard deviation of 8.7 inches.

To find the probability in these scenarios, we need to use the standard normal distribution table, also known as the Z-table, to convert the given values into Z-scores.

A) To find the probability that a randomly chosen child has a height of less than 33.85 inches, we first convert this value to a Z-score using the formula:

Z = (X - μ) / σ

Where:
X = the given value (33.85 inches)
μ = the mean (54.8 inches)
σ = the standard deviation (8.7 inches)

Plugging in the values:

Z = (33.85 - 54.8) / 8.7

Z ≈ -2.40

Now, we need to find the corresponding area under the standard normal curve for a Z-score of -2.40. Looking at the Z-table, we find that the area to the left of this Z-score is approximately 0.0082.

Therefore, the probability that a randomly chosen child has a height of less than 33.85 inches is approximately 0.0082.

B) To find the probability that a randomly chosen child has a height of more than 61 inches, we again convert this value to a Z-score:

Z = (X - μ) / σ

Plugging in the values:

Z = (61 - 54.8) / 8.7

Z ≈ 0.72

Now, we need to find the area to the right of this Z-score. Since the Z-table only provides values for the left side, we need to subtract the area to the left of the Z-score from 1 to get the area to the right.

Using the Z-table, we find that the area to the left of a Z-score of 0.72 is approximately 0.7642. Subtracting this value from 1, we get 1 - 0.7642 = 0.2358.

Therefore, the probability that a randomly chosen child has a height of more than 61 inches is approximately 0.2358.