The length of time it takes college students to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute. Find the cut-off time which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot.

Answer:

Given:

The population mean $(\mu )=3.5$

The population standard deviation $(\sigma )=1$

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,

$$\therefore P(x>x_0)=0.7580$$ $$\therefore 1–P(x>x_0)=1–0.7580$$ $$\therefore P(x<x_0)=0.2420$$ $$\therefore P(z<\frac{x_0–\mu }{\sigma })=0.2420$$ $$\therefore \frac{x_0–\mu }{\sigma }=-0.70$$ $$\therefore \frac{x_0–3.5}{1}=-0.70$$ $$\therefore x_0–3.5=-0.7\times 1$$ $$\therefore x_0=2.80$$