The time a typical American teenager spends doing homework each week is normally distributed with a mean of 4.2 hours and standard deviation of 0.9 hours.
1 Use the Complement Rule : What radio of teenager spend more than 2.4 hours doing homework?
2 Use the Empirical Rule: What is the probability that a randomly selected teenager spends less than 5.1 hours doing homework.
1 Use the Complement Rule : What radio of teenager spend more than 2.4 hours doing homework?
2 Use the Empirical Rule: What is the probability that a randomly selected teenager spends less than 5.1 hours doing homework.
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1)
Did you mean "What proportion of teenagers spend more than 2.4 hours doing homework?"
μ = 4.2
σ = 0.9
standardize x to z = (x - μ) / σ
P(x > 2.4) = P( z > (2.4-4.2) / 0.9)
= P(z > -2) =
= 1-P( z <= -2) = 1-0.0228 = 0.9772
(From Normal probability table)
2)
μ = 4.2
σ = 0.9
standardize x to z = (x - μ) / σ
P(x < 5.1) = P( z < (5.1-4.2) / 0.9)
= P(z < 1) =
(From Normal probability table)
The empirical rule states that 68% of the values lie within 1 standard deviation of the mean.
This means 100-68 = 32% lie above + below
68+(1/2)(32) lie below 1 standard deviation of the mean = 68+16 = 84%
probability = 0.84
Did you mean "What proportion of teenagers spend more than 2.4 hours doing homework?"
μ = 4.2
σ = 0.9
standardize x to z = (x - μ) / σ
P(x > 2.4) = P( z > (2.4-4.2) / 0.9)
= P(z > -2) =
= 1-P( z <= -2) = 1-0.0228 = 0.9772
(From Normal probability table)
2)
μ = 4.2
σ = 0.9
standardize x to z = (x - μ) / σ
P(x < 5.1) = P( z < (5.1-4.2) / 0.9)
= P(z < 1) =
(From Normal probability table)
The empirical rule states that 68% of the values lie within 1 standard deviation of the mean.
This means 100-68 = 32% lie above + below
68+(1/2)(32) lie below 1 standard deviation of the mean = 68+16 = 84%
probability = 0.84