Module 6: Normal life of Normal Distributions

 Hi everyone! 

For this week's assignment, we are looking into normal distributions which is the most common type of distribution in the world of statistics. As a continuous distribution it never fails to be anything short of a bell curve with standard normal distributions having a mean of 0 and a standard deviation of 1.

A.
Consider a population consisting of the following values, which represents the number of ice cream purchases during the academic year for each of the five housemates.
8, 14, 16, 10, 11

a. Compute the mean of this population.




b. Select a random sample of size 2 out of the five members. 
first

Sample is 16 and 8

c. Compute the mean and standard deviation of your sample.






d. Compare the Mean and Standard deviation of your sample to the entire population of this set (8,14, 16, 10, 11).

The mean of the population is 11.8, while the sample mean is 12. The standard deviation of the population is 2.856, while the sample standard deviation is 5.656. The spread is greater for the sample and the mean is relatively the same.



I created a function sd_pop() to find the standard deviation for the population because sd() uses n-1 which is for sample standard deviations.

B. 

Suppose that the sample size n = 100 and the population proportion p = 0.95.

  1. Does the sample proportion p have approximately a normal distribution? Explain.
np >= 10                       n(1-p) >= 10
100 * 0.95 = 95            100(1-.95) = 100 * 0.5 = 5
95 >= 10                      5 is NOT >= 10

No, this sample proportion does not have an approximately normal distribution because n(1-p) used to calculate standard deviation will be too low compared to the mean to have a normal bell shaped distribution.

2. What is the smallest value of n for which the sampling distribution of p is approximately normal? 
200 because it satisfies n(.95) >= 10 and n(.05) >= 10.

The sample mean  from a group of observations is an estimate of the population mean μ . Given a sample of size n, consider n independent random variables X1, X2, ..., Xn, each corresponding to one randomly selected observation. Each of these variables has the distribution of the population, with mean μ  and standard deviation σ .

A. Population mean= (8­­+14+16+10+11)/ 5 = 11.8
B. Sample of size n= 5
C. Mean of sample distribution: 11.8
sample 1=
sample 2=
sample 3 and so on and so forth…
And Standard Error 
Qm=Q/square root of n
=4.4/square root of 5
= 1.96
D. I am looking for table with the following variables X, x=u, and 
(x-u)^2

X (population values)

X - u

(X - u)^2

8

8 - 11.8 = - 3.8

14.44

14

14 - 11.8 = 2.2

4.84

16

16 - 11.8 = 4.2

17.64

10

10-11.8 = -1.8

3.24

11

11- 11.8 = -0.8

0.64


Sum of last column = 40.8

Standard deviation of sample distribution = sqrt(40.8 / (5-1)) = 3.193

C.

From our textbook, Chapter 2 Probability Exercises # 2.4
Simulated coin tossing is probability better done using function called rbinom than using function called sample.  Explain.

The function rbinom() requires a number of observations, number of trials, and the probability of success which in a coin toss matches the characteristics of a binomial distribution with only two mutually exclusive outcomes of heads or tails with identical trials. If we used the sample function there would be a needed vector x of values and the sample size. This function does not specify a need for probability of success and does not require mutually exclusive outcomes of pass or fail making it difficult to use with coin tossing. 

-Ramya's POV


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