Question

The number of passengers who arrive at the platform of a subway station for the 10 am train is a random variable with a mean of 120 and a variance of 16. Find the lower bound of the probability that there will be between 100 and 140 passengers (round off to second decimal place).

Answer 1

Answer:

The lower bound of the probability that there will be between 100 and 140 passengers 96%.

Step-by-step explanation:

The Chebyshev's theorem states that, the probability that X is within k standard deviation of mean is given by,

                                 $P(|X-\mu|<k\sigma)\leq 1-\frac{1}{k^{2}}$

Here, k is any positive number.

The given information is:

Mean (µ) = 120

Standard deviation (σ) = 16.

Then, we need to compute, P (100 ≤ X ≤ 140).

Then,  

$\frac{Upper-\mu}{\sigma}=\frac{140-120}{\sqrt{16}}=5$    

$\frac{\mu-Lower}{\sigma}=\frac{120-100}{\sqrt{16}}=5$              

Since, these two values coincide it implies that the event (66000, 78000) is centered about the mean.

Also, the event P (100 ≤ X ≤ 140) is equivalent to having X within 5 standard deviation of the mean.  

That is, k = 5.

Then for k = 5,  

$1-\frac{1}{k^{2}}=1-\frac{1}{25}=\frac{24}{25}=0.96$

That is, P (100 ≤ X ≤ 140) ≤ 0.96 .

Thus, the lower bound of the probability that there will be between 100 and 140 passengers 96%.