Answer: The lower bound is 37.124 years and upper bound is 108.676 years.
Explanation:
Since we have given that
Mean = 72.90
Standard deviation = 8
At least 95% of population is included.
N = 100
Using Chebyshev's theorem, we have
[tex]100(1-\dfrac{1}{t^2})\%=95\\\\(1-\dfrac{1}{t^2})=0.95\\\\\dfrac{1}{t^2}=0.05\\\\t^2=\dfrac{1}{0.05}=20\\\\t=4.472[/tex]
So, the interval would be
[tex](\mu-t\sigma,\mu+t\sigma)\\\\=(72.90-4.472\times 8,72.90+4.472\times 8)\\\\=(37.124,108.676)[/tex]
Hence, the lower bound is 37.124 years and upper bound is 108.676 years.
