# Solved (Free): For a request to use the campus computing cluster, and knowing that your jobs’ durations are normally distributed with a mean of one hour and a standard

#### ByDr. Raju Chaudhari

Apr 16, 2021

For a request to use the campus computing cluster, and knowing that your jobs' durations are normally distributed with a mean of one hour and a standard deviation of 10 minutes, answer the following inquiries.
(a) What is the probability that the total duration of your 32 jobs is more than 34 hours?
(b) What is the probability that the average duration of your 28 jobs is less than 57 minutes?

#### Solution

Let $X$ denote the jobs duration. Given that $X\sim N(60, 10^2)$.

That is $\mu = 60$ minutes and $\sigma = 10$ minutes.

a. A sample size is $n= 32$.

Then $\sum_{i=1}^{32} Xi \sim N(n\mu, n\sigma^2)$. That is $\sum{i=1}^{32} X_i\sim N(1920, 3200)$.

The probability that the total duration of your 32 jobs is more than 34 hours (i.e. $34*60 = 2040$ minutes)
is

 \begin{aligned} P(\sum_{i=1}^{32} X_i > 2040) &=P\bigg(\frac{\sum_{i=1}^{32} X_i-n*\mu}{\sqrt{n}\sigma} > \frac{2040-1920}{56.5685425}\bigg)\\ &=P\bigg(Z > 2.12\bigg)\\ &= P(Z > 2.12)\\ &=0.017 \end{aligned}

b. A sample size is $n = 28$.

Then $\overline{X}\sim N(\mu, \sigma^2/n)$. That is $\overline{X}\sim N(60, 1.8898^2)$

The probability that the average duration of your 28 jobs is less than 57 minutes is

 \begin{aligned} P(\overline{X} < 57) &=P\bigg(\frac{\overline{X}-\mu}{\sigma/\sqrt{n}} < \frac{57-60}{10/\sqrt{28}}\bigg)\\ &=P\bigg(Z < -1.59\bigg)\\ &= P(Z < -1.59)\\ &=0.0559 \end{aligned}