Zahra B. answered • 11/01/23
Associate Professor of Biostatistics with 10+ years of Experience
To find the probability that a patient will have an age of conception reported between 30 and 32 years old, you can use the z-score formula and the standard normal distribution table.
The z-score formula is: z=x−μ/σ z=σx−μ Where:
- x is the value (age) you want to find the probability for.
- μ is the mean age (27.8 years).
- σ is the standard deviation (2.6 years).
First, find the z-scores for 30 and 32:
For x=30: z1=30−27.8/2.6=0.8462
For x=32: z2=32−27.8/2.6=1.6154
Now, you can look up the probabilities associated with these z-scores in the standard normal distribution table. You'll find the area between these z-scores to get the probability.
P(30≤X≤32)=P(0.8462≤Z≤1.6154)
You can use a standard normal distribution table or a calculator with a cumulative distribution function (CDF) for the standard normal distribution to find this probability.
B) To find the age at conception that corresponds to the lowest 1% of the distribution, you're essentially looking for the 1st percentile of the distribution. You can use the z-score formula again, but this time, you're solving for xx (the age at conception) with a known z-score that corresponds to the 1st percentile.
The z-score associated with the 1st percentile can be found from the standard normal distribution table. It's approximately -2.33 (for a cumulative probability of 0.01).
Now, use the z-score formula to find xx:
−2.33=x−27.8/2.6
Solve for xx:
x=−2.33∗2.6+27.8
Calculate this to find the age at conception corresponding to the lowest 1% of the distribution.
