Answer:
1. a) (1-1) (1-2) (1-3) (2-1) (2-2) (3-1)
2. a) (5-1) (5-2) (5-3) (5-4) (5-5) (5-6)
3. d) (1-3) (2-3) (3-1) (3-2) (3-4) (3-5) (3-6) (4-3) (5-3) (6-3)
Step-by-step explanation:
1. added the numbers in the parenthesis and A had all of the sums less than 5
2. A had all the 5s in the first spot
3. D had the 3s in each parenthesis
Answer:
a) ![\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)
b) ![\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)
Step-by-step explanation:
Given:
The lifetimes of the individual items are independent exponential random variables.
Mean = 200 hours.
Assume, Ti be the time between (
)st and the
failures. Then, the
are independent with
being exponential with rate
Therefore,
a) ![E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]](https://tex.z-dn.net/?f=E%5BT%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20E%5Cleft%5B%5Ctau_%7Bi%7D%5Cright%5D)

![\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)

The variance is given by, ![\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20%5Cmathrm%7BVar%7D%5BT%5D)
![\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)
Answer:
a) The expected number of smokers in a random sample of 140 students from this university is 16.8 smokers.
b) No, it is unlikely that smoking habits and waking up early to go to the gym on Saturday are independent.
Step-by-step explanation:
To calculate the expected numbers of smokers in a sample with size n=140 and proportion p=12%, we use the expected value of the binomial distribution:

The expected number of smokers in a random sample of 140 students from this university is 16.8 smokers.
If we take a sample at the opening of the gym, the sample is not expected to be representative of the population of the students. Most of the students that go to the gym usually have healthy habits, so the proportions of smokers is expected to be lower than the average of the university population.
Answer:
n = 220.1
Step-by-step explanation:
- 4n = 880.4
- n = 880.4 ÷ 4
- n = 220.1
the value of n is 220.1
Answer:
8.72 • 10^7
Step-by-step explanation: