The amount of time Alice spends on her research work every week is normally distributed with a mean of 15 hours with a standard deviation of 10 hours. Her supervisor will get mad at her if she does less than 10 hours of work in any given week. Alice wants to keep her supervisor happy 90% of the time and plans on doing so by changing the standard deviation of her working time. What standard deviation must she achieve?

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The problem describes a normally distributed amount of weekly work time X with mean μ = 15 hours and standard deviation σ, where X ~ N(15, σ^2). The supervisor gets angry if X < 10 hours in a week, and Alice wants this to happen no more than 10% of the time, i.e., P(X ≥ 10) = 0.90.

First, convert the threshold to a standard normal probability. Compute the z-score for X = 10: z = (10 − μ) / σ = (10 − 15) / σ = −5/σ.

We want P(X ≥ 10) = 0.90, which is equivalent to P(Z ≥ −5/σ) = 0.90 for Z ~ N(0,1). Because the standard normal distribution is symmetric, P(Z ≥ −a) = P(Z ≤ a). Therefore, P(Z ≤ −5/σ) = 0.10.

The 10th percentile of the standard normal distribution is approximately z ≈ −1.2816. Thus, −5/σ = −1.2816, which leads to σ = 5 / 1.2816 ≈ 3.90.

Rounding considerations aside, the necessary standard deviation is about 3.90 hours. This ensures that the probability of Alice working fewer than 10 hours in a week is 10%, hence she meets the 90% target.

If you were to check the calculation in a different way, you could set up the inequality P(X < 10) ≤ 0.10, rewrite it with Z, and solve for σ in the same manner, yielding the same result around 3.90 hours.

STA220H1 F LEC0101 Quiz 4

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