The lengths of pregnancies in a small rural village are Normally distributed with a mean of 268 days and a standard deviation of 12 days. What percentage of pregnancies last beyond 290 days?
http://davidmlane.com/hyperstat/z_table.html
To find the percentage of pregnancies that last beyond 290 days, we can use the properties of the normal distribution.
Step 1: Calculate the z-score
The z-score formula is given by:
z = (x - μ) / σ
Where:
x = the value of interest (290 days in this case)
μ = the mean of the distribution (268 days)
σ = the standard deviation of the distribution (12 days)
Plugging in the values:
z = (290 - 268) / 12
z = 22 / 12
z ≈ 1.83
Step 2: Find the percentage using the z-score
To find the percentage beyond a certain z-score, we can look it up in the standard normal distribution table (also known as Z-table).
Looking up the z-score of 1.83 in the table, we find that the area to the left of this z-score is approximately 0.9664.
Step 3: Calculate the percentage beyond 290 days
Since we are looking for the area beyond 290 days, we need to subtract the area to the left of 290 days from 1.
Percentage beyond 290 days = 1 - 0.9664
Percentage beyond 290 days ≈ 0.0336
Therefore, approximately 3.36% of pregnancies in this small rural village last beyond 290 days.