Answer:
![\bar X = (1435040)/(80)=17938](https://img.qammunity.org/2021/formulas/mathematics/college/uwnoa4k0nf36pdlgpr5dz5y0wyfliimhh5.png)
Explanation:
Since we have a groued data for this case we can construct the following table to find the expected value.
Interval Frequency(fi) Midpoint(xi) xi*fi
5001-10000 16 7500.5 120008
10001-15000 14 12500.5 175007
15001-20000 15 17500.5 262507.5
20001-25000 17 22500.5 382508.5
25001-30000 18 27500.5 495009
Total 80 1435040
And we can calculate the mean with the following formula:
![\bar X = (\sum_(i=1)^n f_i x_i)/(n)](https://img.qammunity.org/2021/formulas/mathematics/college/fx8os1ed9fmerw5l53aqbekf32jp67lbyt.png)
Where
![n=\sum_(i=1)^n f_i = 80](https://img.qammunity.org/2021/formulas/mathematics/college/b4wljdlhqbau0rn2d0okbpasnbuysslrmf.png)
And if we replace we got:
![\bar X = (1435040)/(80)=17938](https://img.qammunity.org/2021/formulas/mathematics/college/uwnoa4k0nf36pdlgpr5dz5y0wyfliimhh5.png)