The z-score formula is:
z = (x - mu) / sigma
where z is the z-score, x is the raw score, mu is the mean, and sigma is the standard deviation.
In this case, we are given that mu = 22.5 pounds and sigma = 2.2 pounds. We can use this information to find the z-score for the lower and upper bounds of the weight range, 19.7 pounds and 24.4 pounds, respectively.
For the lower bound of the weight range, we have:
z = (19.7 - 22.5) / 2.2
= -2.8 / 2.2
= -1.2727
For the upper bound of the weight range, we have:
z = (24.4 - 22.5) / 2.2
= 1.9 / 2.2
= 0.8636
Now we can use the standard normal distribution to find the percentage of baby boys who have a z-score between -1.2727 and 0.8636. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1, so the percentage of baby boys who fall within the given weight range is the same as the percentage of baby boys who have a z-score within this range.
To find the percentage of baby boys who have a z-score between -1.2727 and 0.8636, we can use a standard normal distribution table to look up the z-scores and find the corresponding probabilities. The standard normal distribution table shows the percentage of values that fall within a given range of z-scores, so to find the percentage of baby boys who have a z-score between -1.2727 and 0.8636, we need to look up the percentage of values that fall within this range.
According to the standard normal distribution table, the percentage of values that fall within the range of z-scores -1.2727 to 0.8636 is approximately 0.84. Therefore, the percentage of 12-month-old baby boys who weigh between 19.7 and 24.4 pounds is approximately 84%, to the nearest tenth.
