Americans spend an average of 3 hours per day online. If the standard deviation is 32 minutes, find the range in which at least 88.89% of the data will lie. Use Chebyshev's theorem. Hint: You should memorize that when k is 2, it is 75% and when k is 3 it is 88.89%. Also convert hours to minutes and state your answer in minutes.

84 minutes to 276 minutes

Step-by-step explanation:

Chebyshev's theorem states that:

At least 3/4(75% ) of data falls within 2 standard deviation from the mean - that means between μ - 2σ and μ + 2σ

At least 8/9(88.89)% of data falls within 3 standard deviations from the mean - between μ – 3σ and μ + 3σ .

Where k > 1 , 1 - 1/k²

k is 2, it is 75% and when k is 3 it is 88.89%

Americans spend an average of 3 hours per day online. If the standard deviation is 32 minutes, find the range in which at least 88.89% of the data will lies

Step 1

Convert 3 hours to minutes

1 hour = 60 minutes

3 hours = x

Cross Multiply

= 3hours × 60 minutes/1 hour

= 180 minutes

We know from Chebyshev's theorem that:

At least 8/9(88.89)% of data falls within 3 standard deviations from the mean - between μ – 3σ and μ + 3σ .

Hence

μ = 180 minutes, σ =32 minutes

μ – 3σ

= 180 - 3(32)

= 180 - 96

= 84 minutes

μ + 3σ

= 180 + 3(32)

= 180 + 96

= 276 minutes

Therefore the range in which at least 88.89% of the data will lie is 84 minutes to 276 minutes

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.