Question

Americans spend an average of 4 hours per day online.If the standard deviation is 26 minutes, find the range in whitch at least 88.89% of the data will lie. Use Chebyshev's theorem. Round your K to the nearest whole number.

Answer 1

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, $\bar x$ = $4$ hrs = $240$ mins

Standard deviation, $s$ = $26$ 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 = $\bar x-ks$

Higher range = $\bar x +ks$

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.

⇒ $88.89\% = 0.8889$

⇒ $1-\frac{1}{k^2} = 0.8889$

⇒ $1-0.8889=\frac{1}{k^2}$

⇒ $0.1111=\frac{1}{k^2}$

⇒ $\frac{1}{0.1111} =k^2$

⇒ $\sqrt{\frac{1}{0.1111} } =k$

⇒ $3=k$

Plugging the value of k=3 in the formula above.

The range is $\bar x-ks,\bar x+ks$ .

⇒ $240-3(26), 240+3(26)$

⇒ $240-78,240+78$

⇒ $162,318$

So,

The range in which at least 88.89% of the data will lie is (162,318).