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 the interval around the mean that contains 95% of the attendance?

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)

incomplete question. how do you expect anyone to you when you do not post the question in its entirety?