Yes he is correct, 100% would be 10 hours again, so 110% would add the extra hour to make it 11 hours
Less because l don't be yelling at me what's wrong wit you.
Answer:
The range in which at least 88.89% of the data will lie is (162,318).
Step-by-step explanation:
Given:
Average time spend online,
=
hrs =
mins
Standard deviation,
=
mins
We have to find the range in which at least 88.89% of the data will lie by using Chebyshev's theorem :
Formula :
Lower range =
Higher range = 
Now we have to find k values.
We know that the percentage of the data that lies within 'k' standard deviation is at least,1-1/k^2 and k > 1.
⇒ 
⇒
⇒
⇒ 
⇒
⇒ 
⇒ 
Plugging the value of k=3 in the formula above.
The range is
.
⇒
⇒
⇒ 
So,
The range in which at least 88.89% of the data will lie is (162,318).