At a customer service call center for a large company, the number of calls received per hour is normally distributed with a mean of 130 calls and a standard deviation of 10 calls. What is the probability that during a given hour of the day there will be less than 129 calls, to the nearest thousandth?

0.460

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 129 calls

μ is the population mean = 130 calls

σ is the population standard deviation = 10 calls

Hence:

Z score = x - μ/σ

=129 - 130/10

=-0.1

Probability value from Z-Table:

For x less than 129 calls = x < 129

P(x<129) = 0.46017

Approximately to the nearest thousandth = 0.460

Therefore, the probability that during a given hour of the day there will be less than 129 calls, to the nearest thousandths is 0.460

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