Answer:
0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.
Step-by-step explanation:
We have the mean during a time-period, which means that the Poisson distribution is used to solve this question.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval.
A service center receives an average of 0.6 customer complaints per hour.
This means that
, in which h is the number of hours.
Determine the probability that exactly four complaints will be received during the next eight hours.
8 hours means that
.
The probability is P(X = 4).


0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.
The 68-95-99.7 rule tells us 68% of the probability is between -1 standard deviation and +1 standard deviation from the mean. So we expect 75% corresponds to slightly more than 1 standard deviation.
Usually the unit normal tables don't report the area between -σ and σ but instead a cumulative probability, the area between -∞ and σ. 75% corresponds to 37.5% in each half so a cumulative probability of 50%+37.5%=87.5%. We look that up in the normal table and get σ=1.15.
So we expect 75% of normally distributed data to fall within μ-1.15σ and μ+1.15σ
That's 288.6 - 1.15(21.2) to 288.6 + 1.15(21.2)
Answer: 264.22 to 312.98