Answer:
Given:
The population mean $(μ) = 3.5$
The population standard deviation $(σ) = 1$
Solution:
→ Let $x_0$ be the cut-off time which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot,
$$\therefore P(x > x_0) = 0.7580$$ $$\therefore 1 – P(x > x_0) = 1 – 0.7580$$ $$\therefore P(x < x_0) = 0.2420$$ $$\therefore P\left( z < \frac{x_0 – \mu}{\sigma} \right) = 0.2420$$ $$\therefore \frac{x_0 – \mu}{\sigma} = -0.70$$ $$\therefore \frac{x_0 – 3.5}{1} = -0.70$$ $$\therefore x_0 – 3.5 = -0.7 \times 1$$ $$\therefore x_0 = 2.80$$
