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Researchers at a computer company conducted a survey on the number of hours computers are used in households each day. The data from a random sample of 2,100 homes produced a normal distribution with:

- A mean length of time: 5.3 hours per day
- A standard deviation: 0.9 hours

Based on the sample, what percentage best represents the households that use a computer less than 5 hours or more than 6 hours per day?

Answer :

The percentage of households using a computer for less than 5 or more than 6 hours per day, based on the given normal distribution, is approximately 58.84%.

Given a normal distribution with a mean (μ) of 5.3 hours and a standard deviation (σ) of 0.9 hours, we can find these percentages using Z-scores and the standard normal distribution table.

Calculate the Z-score for 5 hours:

(Z = (X - μ) / σ)

Z5 = (5 - 5.3) / 0.9 = -0.33

Calculate the Z-score for 6 hours:

(Z = (X - μ) / σ)

Z6 = (6 - 5.3) / 0.9 = 0.78

Find the cumulative probability for these Z-scores:

P(Z < -0.33) ≈ 0.3707P(Z < 0.78) ≈ 0.7823

Calculate the percentage of households using computers for

more than 6 hours per day:

P(X > 6) = 1 - P(Z < 0.78) = 1 - 0.7823 = 0.2177 (or 21.77%)

Calculate the percentage of households using computers for

less than 5 hours per day:

P(X < 5) = P(Z < -0.33) = 0.3707 (or 37.07%)

Total percentage of households using a computer

less than 5 hours or more than 6 hours per day:

P(X < 5 or X > 6) = P(X < 5) + P(X > 6) = 0.3707 + 0.2177 = 0.5884 (or 58.84%)

Therefore, the percentage of all households that use a computer
less than 5 hours or more than 6 hours per day is approximately 58.84%.

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