Evaluate the limit as h goes to 0 of (3+h)^4 -81/h
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Given:
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).
μ = 500 days, the population mean
σ = 60 days, the population standard deviation
Therefore
μ + σ = 560
μ - σ = 440
μ + 2σ = 620
μ - 2σ = 380
μ + 3σ = 680
μ - 3σ = 320
The figure shown below illustrates the normal distribution
About 68% of the total area lies in x = (μ-σ, μ+σ)
About 95% of the total area lies in x = (μ-2σ, μ+2σ)
About 99.7% of the total area lies in x = (μ-3σ, μ+3σ).