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, [tex]\bar x[/tex] = [tex]4[/tex] hrs = [tex]240[/tex] mins
Standard deviation, [tex]s[/tex] = [tex]26[/tex] 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 = [tex]\bar x-ks[/tex]
Higher range = [tex]\bar x +ks[/tex]
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.
⇒ [tex]88.89\% = 0.8889[/tex]
⇒ [tex]1-\frac{1}{k^2} = 0.8889[/tex]
⇒ [tex]1-0.8889=\frac{1}{k^2}[/tex]
⇒ [tex]0.1111=\frac{1}{k^2}[/tex]
⇒ [tex]\frac{1}{0.1111} =k^2[/tex]
⇒ [tex]\sqrt{\frac{1}{0.1111} } =k[/tex]
⇒ [tex]3=k[/tex]
Plugging the value of k=3 in the formula above.
The range is [tex]\bar x-ks,\bar x+ks[/tex] .
⇒ [tex]240-3(26), 240+3(26)[/tex]
⇒ [tex]240-78,240+78[/tex]
⇒ [tex]162,318[/tex]
So,
The range in which at least 88.89% of the data will lie is (162,318).
