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).

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,

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,  

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,  

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%.

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!