Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.
- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )
E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
Explanation + answers
Because there are lengths going from 0 to 1, the lines must mean either decimals or fractions (we'll use fractions for this.)
1. There are twelve lines from 0 and 1, which we can use as the denominator for our fraction. This means the length of each line is 1/12.
2. In order to find where K's point is at, we simply need to count until we get to it. After counting, I see that K is on point 8/12, which we can simplify to get a smaller number. If we simplify once, we get 4/6, which we can again simplify to get 2/3. This gives us the answer K is on point 8/12 or 2/3.
Standard form is ax^2+bx+c=0
First evaluate the exponent and expand (foil).
y= 3(x^2-4x+4)+1
Then distribute the 3
y= 3x^2-12x+12+1
Combine like terms
y=3x^2-12x+13
Final answer: 3x^2-12x+13=0
Answer:
she will cover a distance of 0.85 km.
Step-by-step explanation:
We have,
Grace swam 0.6 kilometres during her first week of training and 0.25 kilometres during her second week of training.
It is required to find total distance covered by her in all.
It can be calculated simply adding the distance covered in first and second week of training. So,
Total distance, D = 0.6 km +0.25 km = 0.85 km
Hence, she will cover a distance of 0.85 km.
Answer:
A student finds a 95% confidence interval of (16.34,18.69) for the mean number of species counted. This is a valid interval because the mean number of species or any population mean does not necessarily have to be a whole number, as stated by the student.
This given confidence interval of (16.34,18.69) helps us to simply estimate the mean species counted.
