Question

time spent online americans spend an average of 3 hours per day online. if the standard deviation is 37 minutes, find the range in which at least 88.89% of the data will lie. use chebyshevs theorem. round your k to the nearest whole number. at least 88.89% of the data will lie between and minutes.

Explanation:

Step1: Use Chebyshev's formula

Chebyshev's formula is \(P(|X - \mu|\leq k\sigma)\geq1-\frac{1}{k^{2}}\). We know that \(P(|X - \mu|\leq k\sigma)= 88.89\%=\frac{8}{9}\). Then \(1-\frac{1}{k^{2}}=\frac{8}{9}\).

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Step2: Calculate the range

The mean \(\mu=3\times60 = 180\) minutes (since 3 hours = 180 minutes), and the standard deviation \(\sigma = 37\) minutes.
The lower bound is \(\mu - k\sigma=180-3\times37=180 - 111=69\)
The upper bound is \(\mu + k\sigma=180 + 3\times37=180+111 = 291\)

Answer:

69 and 291