Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Ask your questions and receive precise answers from experienced professionals across different disciplines. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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
#SPJ4
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
