Answer:
Step-by-step explanation:
HEADS
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x3
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
C) 1.94
A) 35%
A) 200%
A) 270
A) 252
