Question

Exercise 8. (10 points] Sampling Distribution The population of students's detention D follows the dustribution: 1 0 0.55 2. 0.15 3 0.05 Tot 1 0.25

a) Compute the sampling distribution for the sample avrega X¯ when n = 2. What is the probability that the sample of 2 students get more than 1 detentions on average?

b) Compute the approximate distribution for a sample of n = 70. What is the probability that the sample of 70 students get more than 1 detentions on average?

Answer 1

a) For a sample size of n = 2, the probabilities here that the sample mean is less than equal to 1 is computed here as:

(0,0): 0.55*0.55 = 0.3025
(0,1): 0.55*0.25 = 0.1375
(1,0): 0.25*0.55 = 0.1375
(1,1): 0.25*0.25 = 0.0625
(0,2): 0.55*0.15 = 0.0825
(2,0):0.15*0.55 = 0.0825

Therefore, P(Sample Mean <= 1) = adding all the above probabilities

= 0.3025 + 0.1375 + 0.1375 + 0.0625 + 0.0825 + 0.0825

= 0.8050

Therefore the probability that the sample mean is more than 1 is computed here as:
= 1 - P( Sample mean <= 1) = 1 - 0.805 = 0.1950

Therefore 0.195 is the required probability here.

b) Here, we first compute the population mean and standard deviation as:

Mean = E(X) = 0*0.55 + 1*0.25 + 2*0.15 + 3*0.05 = 0.7

E(X²) = 0²*0.55 + 1²*0.25 + 2²*0.15 + 3²*0.05 = 1.3

Therefore, Var(X) = E(X²) - [E(X)]² = 1.3 - 0.7² = 0.81

Therefore using central limit theorem the distribution of sample size here is computed as:

$\bar X \sim N(\mu = 0.7, \sigma^2 = \frac{0.81}{n})$

$\bar X \sim N(\mu = 0.7, \sigma^2 = \frac{0.81}{70})$

$\bar X \sim N(\mu = 0.7, \sigma^2 = 0.01157)$

Therefore the probability now here is computed as:

P(X > 1)

Converting it to a standard normal variable, we have here:

1- 0.7 P(Z > 0.01157

Getting it from the standard normal tables, we have here:

Therefore 0.0026 is the required probability here.