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Answer:
The interval that contains 95% of the attendance is (23099.6, 60104.4)
Step-by-step explanation:
We are given that average number of fans per game that is mean=41,602 and the attendance follows the normal probability distribution with standard deviation of 9440 per game.
And we are asked to find the interval around the mean that contains 95% of the distance.
Let us find z-value for 95% confidence interval that is for 0.95 and it is z=1.96.
Hence interval is (mean - (z value)X(standard deviation), mean+(z value)X(standard deviation)
that is (41602-1.96X9440, 41602+1.96X9440)
(23099.6, 60104.4)
The interval around the mean that contains 95% of the attendance is 56706.
Given
During a particular baseball season, a team averaged 41,602 fans per home game.
suppose attendance during the season follows the normal probability distribution with a standard deviation of 9,440 per game.
What is a z-score?
A z score helps to calculate the probability of a score occurring within a standard normal distribution.
The z-score is calculated using the following formula;
[tex]\rm z-score=\dfrac{x-\mu}{\sigma}[/tex]
Where, μ = 41,602, the mean and σ = 9,440, the standard deviation.
Let us find the z-value for the 95% confidence interval that is for 0.95 and it is z =1.96.
Therefore,
The interval around the mean that contains 95% of the attendance is;
[tex]\rm z-score=\dfrac{x-\mu}{\sigma}\\\\1.6=\dfrac{x-41,602}{9440}\\\\1.6\times 9440=x-41602\\\\x-41602=15104\\\\x=15104+41602\\\\x=56706[/tex]
Hence, the interval around the mean that contains 95% of the attendance is 56706.
To know more about z-score click the link is given below.
https://brainly.com/question/10801289
