Question

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. (ii) Find the probability that Paulo’s lunch break lasts for between 40 and 46 minutes on every one of the next four days.

Answer 1

Let $X$ denote the length of Paulo's lunch break.

We are given that, $\sigma=5$ and $p(x\gt 52)={1\over4}$

$a)$

To solve for $\mu$, we standardize $p(x\gt 52)={1\over4}$ as follows.

$p(x\gt 52)=p(Z\gt {52-\mu\over \sigma})=p(Z\gt {52-\mu\over 5})={1\over4}$

So,

$p(Z\lt {52-\mu\over 5})=1- p(Z\gt {52-\mu\over 5})=1-{1\over4}={3\over4}$. This shows that,

$\phi({52-\mu\over 5})={3\over4}$, implying that ${52-\mu\over 5}=Z_{1-{3\over4}}=Z_{1\over 4}$ where $Z_{1\over 4}$ is the standard normal table value that leaves an area of ${1\over4}$ to the right and an area of ${3\over4}$ to the left. From the normal tables, $Z_{1\over4}=0.6744898$.

Thus,

${52-\mu\over5}=0.6744898\implies 52-\mu=3.372449\implies \mu=48.627551$.

b$)$

Here we determine the probability, $p(40\lt x\lt 46)$. So, $p(40\lt x\lt 46)=p({40-48.63\over5}\lt Z\lt {46-48.63\over5})=p(-1.73\lt Z\lt-0.53)=\phi(-0.53)-\phi(-1.73)=0.29806-0.04182=0.25624$

The probability that Paulo’s lunch break lasts for between 40 and 46 minutes on every one of the next four days is 0.25624.