It is not surprising that babies with low birth weights are at risk for developmental difficulties, but extraoridinarily large babies also face a higher than noraml risk of medical problems. In the United States the average full term single birth baby has a weight of 3.4 kg with a standard deviation of 0.6 kg.
a) Babies below 2.5 kg in weight are considered to be high risk/low weight deliveries. Assuming birth weghts are normally distributed, what percentage of births would be in this category?
b) Babies above 4.6 in weight are considered to be high risk/high birth weight deliveries. What percentage of births would fall in this category?
c) Suppose a new study claims that only the middle 80% of the birth weights should be considered normal. What would be the new cut-off points for low and high weight risk deliveries?
****PLEASE provide steps & answers! Thanks :)
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability for each Z score.
For C, use P(±.40) to find Z and insert into equation above.
To find the percentage of births falling in a certain weight category or range, we need to use the concept of z-scores and the standard normal distribution. The z-score measures the number of standard deviations a particular value is away from the mean.
a) To find the percentage of births below 2.5 kg (low weight deliveries), we first need to calculate the z-score for this weight using the formula:
z = (x - μ) / σ
where:
x = the specific value (2.5 kg)
μ = the mean (3.4 kg)
σ = the standard deviation (0.6 kg)
Substituting the values, we get:
z = (2.5 - 3.4) / 0.6
z = -0.9 / 0.6
z = -1.5
Now, we need to find the area under the standard normal distribution curve to the left of z = -1.5. This can be done using statistical tables or software.
Looking up the z-score -1.5 in the standard normal distribution table, we find that the area to the left of it is approximately 0.0668 or 6.68%.
Therefore, approximately 6.68% of births would be categorized as high-risk/low-weight deliveries.
b) To find the percentage of births above 4.6 kg (high birth weight deliveries), we follow a similar process.
z = (4.6 - 3.4) / 0.6
z = 1.2 / 0.6
z = 2.0
Looking up the z-score 2.0 in the standard normal distribution table, we find that the area to the left of it is approximately 0.9772 or 97.72%.
Since we want the percentage of births above 4.6 kg, we subtract the area to the left from 1:
1 - 0.9772 = 0.0228 or 2.28%
Therefore, approximately 2.28% of births would be categorized as high-risk/high-weight deliveries.
c) If only the middle 80% of birth weights are considered normal, we need to find the new cutoff points for low and high weight risk deliveries.
Since the middle 80% refers to the range that covers 80% of the data, we need to find the z-scores that correspond to the upper and lower 10th percentiles of the standard normal distribution.
The lower 10th percentile corresponds to an area to the left of z = -1.28, while the upper 10th percentile corresponds to an area to the right of z = 1.28.
Using the z-score formula, we can solve for the new cutoff points:
-1.28 = (x - 3.4) / 0.6 (lower cutoff)
1.28 = (x - 3.4) / 0.6 (upper cutoff)
Solving for 'x' in both equations:
x - 3.4 = -1.28 * 0.6
x - 3.4 = -0.768
x = 2.632
x - 3.4 = 1.28 * 0.6
x - 3.4 = 0.768
x = 4.168
The new cutoff points for low and high weight risk deliveries would be approximately 2.632 kg and 4.168 kg, respectively.