Answer:
a) To find the probability that a randomly chosen 12-month-old baby girl weighs less than 19.98 pounds, we need to standardize the weight using the z-score formula:
z = (x - μ) / σ
where x is the weight we want to find the probability for, μ is the mean weight, and σ is the standard deviation.
Substituting the given values, we get:
z = (19.98 - 21) / 2.18
z = -0.46
Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -0.46 is approximately 0.3222.
Therefore, the probability that a randomly chosen 12-month-old baby girl weighs less than 19.98 pounds is about 0.3222 or 32.22%.
b) To find the probability that a randomly chosen 12-month-old baby girl weighs more than 22.84 pounds, we again need to standardize the weight using the z-score formula:
z = (x - μ) / σ
Substituting the given values, we get:
z = (22.84 - 21) / 2.18
z = 0.84
Using a standard normal distribution table or calculator, we find that the probability of a z-score greater than 0.84 is approximately 0.2005.
Therefore, the probability that a randomly chosen 12-month-old baby girl weighs more than 22.84 pounds is about 0.2005 or 20.05%.
c) To find the proportion of 12-month-old baby girls who weigh between 19.36 and 22.08 pounds, we need to standardize both weights using the z-score formula:
z1 = (19.36 - 21) / 2.18
z1 = -0.73
z2 = (22.08 - 21) / 2.18
z2 = 0.53
Using a standard normal distribution table or calculator, we find the area under the curve between these two z-scores, which is approximately 0.5722.
Therefore, the proportion of 12-month-old baby girls who weigh between 19.36 and 22.08 pounds is about 0.5722 or 57.22%.