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Sagot :
The average customers are spending in the range of ( 196.36 , 203.64) for the given mean and standard deviation.
As given in the question,
Sample size 'n' = 50
Sample mean 'μ ' is equal to 200
Standard deviation 'σ' = 10
z-score for 99% Confidence interval = 2.576
Confidence interval = mean ± z ( σ / √n )
Substitute the values we get,
Confidence Interval = 200 ± 2.576( 10 / √50)
= 200± 2.576 ( 1.414)
= 200 ± 3.642464
Lower bound of the required confidence interval = 200 - 3.642464
= 196.36
Upper bound of the required confidence interval = 200 + 3.642464
= 203.64
Therefore, the average customer is spending between ( 196.36 , 203.64) range from the given mean and standard deviation.
Learn more about standard deviation here
brainly.com/question/23907081
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