Answer:
0.0035289
Step-by-step explanation:
From the question;
mean annual salary = $63,500
n = sample size = 31
Standard deviation = $6,200
Firstly, we calculate the z-score of $60,500
Mathematically;
z-score = x-mean/SD/√n = (60500-63500)/6200/√(31) = -2.6941
So we want to find the probability that P(z < -2.6941)
We can get this from the standard normal table
P( z < -2.6941) = 0.0035289
The answer to your question is 215
I’m pretty sure it’s y if not I’m sorry
218, 314 are the two values do the middle 95% of the lengths of all pregnancies fall.
