The length of Paulo’s lunch break follows a normal distribution with mean u and standard deviation 5 minutes. On one day in four, on average, his lunch break lasts for more than 52 minutes. (i) Find the value of u.

Let $X=$ the length of Paulo’s lunch break in minutes: $X\sim N(u, \sigma^2).$

Given $\sigma=5\ min, P(X>52)=\dfrac{1}{4}=0.25$

$P(X>52)=1-P(X\le 52)$

$=1-P(Z\le \dfrac{52-u}{5})=0.25$

$P(Z\le \dfrac{52-\mu}{5})=0.75$

$\dfrac{52-u}{5}\approx0.6745$

$u=52-5(0.6745)$

$u=48.628\ min$