Question

Think of some discrete random variable you observe on a regular basis. For example, it could be the (rounded) number of hours you sleep, how many gallons of gas are in your car when you get into it, how many boxes of cereal are in your house, how many days between grocery shopping, etc. (just make sure it takes only integer values). Try to list all of the possible values that this discrete random variable can take. If you can, collect some frequency data – give the relative frequency table and use this as an estimate of the probability distribution. Calculate the expected value and the standard deviation for this probability distribution. Interpret these parameters, and discuss whether they make sense based on your experience. Week 4 Responses (100+ words, x2): Look at your classmates’ distribution. Is there any well-known distribution that could be used to model their random phenomenon? Some well-known discrete distributions are: Uniform, Bernoulli, Binomial, Geometric, and Poisson (but there are othbe appropriate. Post a picture of tdiscrete distribution and a histogram of the frequency data from the original post, and comment on what is similar and different. Are there any outliers that, if removed, would make the frequency data match the distribution really well?

Answer 1

Sol:

Discrete random variable: "Number of days I eat something outside in a given month".

Possible values of the above random variable: 0, 1, 2, 3, 4,............, 29, 30, 31

Data:

I shall take one year (12 months) data.

0, 5, 7, 1, 3, 5, 4, 10, 8, 0, 2, 3

Relative frequency table:

No. of days	Mid value	Frequency	Relative frequency
0 - 2	1	4	4/12 =0.3333
3 - 5	4	5	5/12 =0.4167
6 - 8	7	2	2/12 =0.1667
9 - 11	10	1	1/12 =0.0833
Total		12	1

Probability distribution:

Mid value: Xi	Probability: Pi	PiXi	Pi[Xi - E(X)]2
1	0.3333	0.3333	0.3333(1-4)2 =2.9997
4	0.4167	1.6668	0.4167(4-4)2 =0
7	0.1667	1.1669	0.1667(7-4)2 =1.5003
10	0.0833	0.8330	0.0833(10-4)2 =2.9988
Total	1	4	7.4988

Expected value, E(X) = =4 days in a month.

Standard deviation, = = =2.74 days in a month. ​​​​​​

Interpretation:

E(X): "On an average, the number of days I eat something outside in a given month is 4 days".

: "On an average, the variation of the number of days that I eat something outside in a given month is 2.74 days from the expected value of 4 days".

I am sure that these parameters make sense based on my experience because it's true that on an average, I eat something outside in a given month for about 4 days and it varies by about 3 days (on an average), i.e., mostly upto 4+3 =7 days and 4-3 =1 day and again less than 4 days is most common than that of more than 4 days and thus, 2.74 which is less than 3 for the standard deviation makes sense.