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Answer:
On the first day, 98.75% probability that the sample average amount of time taken is at most 11 min.
On the second day, 99.29% probability that the sample average amount of time taken is at most 11 min.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
First day
The probability is the pvalue of Z when X = 11. So
By the Central Limit Theorem
has a pvalue of 0.9875
On the first day, 98.75% probability that the sample average amount of time taken is at most 11 min.
Second day
The probability is the pvalue of Z when X = 11. So
By the Central Limit Theorem
has a pvalue of 0.9929
On the second day, 99.29% probability that the sample average amount of time taken is at most 11 min.
The various answers to the question are:
- To answer 90% of calls instantly, the organization needs four extension lines.
- The average number of extension lines that will be busy is Four
- For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.
a)
A number of extension lines needed to accommodate $90 in calls immediately:
Use the calculation for busy k servers.
The probability that 2 servers are busy:
The likelihood that 2 servers will be busy may be calculated using the formula below.
Hence, two lines are insufficient.
The probability that 3 servers are busy:
Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.
Thus, three lines are insufficient.
The probability that 4 servers are busy:
Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.
Generally, the equation for is mathematically given as
To answer 90% of calls instantly, the organization needs four extension lines.
b)
The probability that a call will receive a busy signal if four extensions lines are used is,
Therefore, the average number of extension lines that will be busy is Four
c)
In conclusion, the Percentage of busy calls for a phone system with two extensions:
The likelihood that 2 servers will be busy may be calculated using the formula below.
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.
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