Question

Answer for part A: I need help with part B

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Part B: Sampling and Random Variable You already have ten marked pennies (ones with numbers from Part A) and 15 unmarked pennies.

Thought experiment: Throw them all in a jar and shake. Without looking, pull three out and record how many of them are marked (have a number). You will get 0, 1, 2, or 3 marked coins.

How many different samples of 3 pennies out of 25 can you get? (Order doesn’t matter.) Answer: 2,300                      Show why 2,300 is the answer.

How many of those samples would have 0, 1, 2, 3 marked coins?

# of marked pennies in sample          0                         455          1                         1,050          2                          675           (Show why this count should be 675.)          3                          120                       Total     2,300       (Notice this total matches the total above.)

If you draw a simple random sample of size 3, each sample is equally likely. Counting the number of marked coins gives us a discrete random variable, X.

P(X=0)=      455/2,300     =    .1978 P(X=1)= P(X=2)= P(X=3)=

Find the rest of these probabilities. Then find the mean and standard deviation of this discrete random variable.

Now, really and truly put the ten marked coins in a jar with 15 unmarked ones. Shake, pull out three without looking. Write down the number of marked coins. Put them all back, shake, draw again, and count marked coins again. Do this a total of 20 times. Now you have 20 pieces of data. Write down the data set. Compute the sample proportions. How do your sample proportions compare to the probabilities you computed above? Find the mean and standard deviation of the data set. Are the expected value and standard deviation of the random variable close to the mean and standard deviation of the data set? Should they be? Why?

Answer 1

Solution PART B

Back-up Theory

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}…...................................…(1)

Values of ⁿC_r can be directly obtained using Excel Function: Math & Trig COMBIN….......................................…. (1a)

If a discrete random variable, X, has probability function, p(x), x = x₁, x₂, …., x_n, then

Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x…..…. (2)

E(X²) = Σ{x².p(x)} summed over all possible values of x…………………………………..(3)

Variance of X = Var(X) = σ² = E(X²) – {E(X)}²……………………………………………..(4)

Standard Deviation of X = SD(X) = σ = sq.rt of Var(X) …………………..………………..(5)

Now, to work out the solution

We have 25 pennies of which 10 are marked pennies and 15 are unmarked pennies .............................

Part (a)

If a sample of 3 pennies has x marked pennies, then it must have (3 - x) unmarked pennies.

So, x marked pennies implies x marked pennies and (3 - x) unmarked pennies.

Hence, vide (1), number of combinations with x marked pennies = (¹⁰C_x) x (¹⁵C_{3- x}) ……….

Using (1a), the above is presented for x = 0, 1, 2 and 3

# of marked pennies	# of unmarked pennies	N(marked)	N(unmarked)	Total
0	3	1	455	455
1	2	10	105	1050
2	1	45	15	675
3	0	120	1	120
			Total	2300

Answer 1

Part (b)

Using the above, probability distribution of X is:

# of marked pennies - x	Probability
0	455/2300 = 0.1978
1	1050/2300 = 0.4565
2	675/2300 = 0.2935
3	120/2300 = 0.0522
Total	1.0000

Answer 2

Part (c)

Vide (2), (3), (4) and (5), the required calculations are given below.

x	p(x)	x.p(x)	x2.p(x)
0	0.1978	0.0000	0.0000
1	0.4565	0.4565	0.4565
2	0.2935	0.5870	1.1739
3	0.0522	0.1565	0.4696
Total	1.0000	1.2000	2.1000
vide (2), Mean = 1.20 Answer 3
Vide (4), Variance = 2.10 - 1.22 = 0.66
Vide (5), standard deviation = 0.81 Answer 4

DONE