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)

from the figure we get,

the x axis shows   the months and,

the y-axis shows the number of satisfied customer.

each line or each cell in upward direction (y-axis) shows 500.

the appropriate scale for the y - axis is: -

a = 2000 customers

b = 4000 customers

c = 6000 customers

d = 8000 customers

e = 10000 customers

f = 12000 customers

g = 14000 customers