Question

Traveling between two campuses of a university in a city via shuttle bus​ takes, on​ average, 28 minutes with a standard deviation of 5 minutes. In a given​ week, a bus transported passengers 100 times. The probability that the average transport time was bigger than 27 minutes is

Answer 1

Answer: 0.9772

Step-by-step explanation:

Let X be a random variable that represents the transport time .

We assume X is normally distributed with a mean $\mu=28\text{ minutes}$ and a standard deviation $\sigma=5\text{ minutes}$.

Sample size : n= 100

The probability that the average transport time was bigger than 27 minutes:

$P(\overline{X}>27)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{27-28}{\dfrac{5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-1}{\dfrac{5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>-2)=P(Z<2)\ \ \ [P(Z>-z)=P(Z<z)]\\\\=0.9772\ \ \ [\text{By p-value table}]$

Hence, the required probability =0.9772