A large surey of customers who called a customer service number found that 46% of customers were satisfied with the service they received. Since the survey, the customer service agents have received additional training. After the training, interviews with 10 randomly selected customers found that 6 of them were satisfied with the service they received.

a) What is the probability that 60 or more out of 100 randomly selected customers would have said th3ey were satisfied with the service they received, BEFORE the training.

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**This is what is in the answer key:**
Normal Approx to Binomial

N = 100

np = 46

n(1-p) = 54

Since these are >5, we can use the Normal Approximation

Mean = 46

SD = 4.983974

z=2.708682

prob from table = 0.4966

(P>60) = 0.0034

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How do they get those last 3 lines? I tried using the z-score formula (x-mean/sd) but it doesn't give me that z score. Can someone help me explain how they get from that z-score to the final answer?

Also what do they mean by : "Since these are >5, we can use the Normal Approximation"?

Thanks