Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.
1 answer:
Answer:
The likelihood that Albert's sample of 64 will have a mean waiting time between 13.5 and 16.5 minutes is 0.9973.
Step-by-step explanation:
Prior concepts
A normal distribution is characterized as a "probability distribution that is symmetric around the mean, indicating that data close to the mean are more frequent than those further away".
The Z-score refers to "a statistical measurement that reflects the relationship of a value to the mean of a group, measured in standard deviations".
Let X denote the random variable of interest, and we identify its distribution:
Also, let signify the sample mean, whose distribution is:
In this case,
Solution to the problem
We seek this probability
Applying the Z-score formula to the probability results in:
To determine these probabilities, we can refer to normal distribution tables, use Excel, or a calculator.
The likelihood that Albert's sample of 64 will have a mean waiting time between 13.5 and 16.5 minutes is 0.9973.
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Response:
24.2 (rounded to the nearest tenth)
Detailed breakdown:
The data type involved is ungrouped
standard deviation =√ [∑ (x-μ)² / n]
average (μ)=∑x/n
=
=33.3
For the values 5, 5, 6, 12, 13, 26, 37, 49, 51, 56, 56, 84, the x-μ results will be
-28.3, -28.3, -27.3, -21.3, -7.3, 3.7, 15.7,17.7, 22.7,22.7,50.7
Calculating (x-μ)² gives: 800.89, 800.89, 745.29, 453.69, 53.29, 13.69, 246.49, 313.29, 515.29, 515.29, 2570.49
Then, ∑ (x-μ)² results in = 7028.59
Thus, standard deviation = √(7028.59 / 12)
=24.2