Statistical Inference I: J. Lee Assignment 2 Problem 1. In ✓ Solved

In your pocket, you have 1 dime, 2 nickels, and 2 pennies. You select 2 coins at random (without replacement). Let X represent the amount (in cents) that you select from your pocket.

(a) Give (explicitly) the probability mass function for X. Also show a plot of it.

(b) Give (explicitly) the cdf, F(x), for X. Also show a plot of it.

(c) How much money do you expect to draw from your pocket?

Problem 2. Consider a random variable X whose distribution function (cdf) is given by FX(x) = 8 >>>>< >>>>: 0 if x < 2 0.1 if 2 ≤ x < 1.1 0.3 if 1.1 ≤ x < 2 0.6 if 2 ≤ x < 3 1 if x ≥ 3.

(a) Give the probability mass function, p(x), of X, explicitly.

(b) Compute P(2 < X < 3).

(c) Compute P(X ≥ 3).

(d) Compute P(X ≥ 3 | X = 0).

(e) What is the cdf (distribution function) of Y = X²? (be explicit!)

(f) Compute E(X). Also compute E(X³ cos(X)).

Problem 3. Six men and five women apply for a job at Alpha, Inc. Three of the applicants are selected (at random) for interviews. Let X denote the number of women in the interview pool.

(a) Give the probability mass function, p(x), of X, explicitly.

(b) Find the probability that either one or two women are in the interview pool.

(c) How many women do you expect to be in the interview pool?

Problem 4. Consider a random variable X whose probability mass function (pmf) is given by p(x) = 8 >>>>>>< >>>>>>: p if x = 1.9 0.1 if x = 0.1 0.3 if x = 20p p if x = 3 4p if x = 4 0 otherwise.

(a) What is p?

(b) Compute P(1.9 ≤ |X| ≤ 3).

(c) What is F(0)? What is F(2)? What is F(F(3.1))?

(d) Sketch a plot of the function F(x). (Make sure to label the coordinates on the axes!)

(e) What is P(2X ≥ 3 | X ≥ 2.0)?

(f) Compute E(X). (g) Compute E(F(X)).

Paper For Above Instructions

Problem 1:

In this problem, we have a collection of coins consisting of 1 dime (10 cents), 2 nickels (5 cents each), and 2 pennies (1 cent each). This gives the following values for X when selecting 2 coins without replacement:

  • Two pennies => 1 + 1 = 2 cents

  • One penny and one nickel => 1 + 5 = 6 cents

  • One penny and one dime => 1 + 10 = 11 cents

  • Two nickels => 5 + 5 = 10 cents

  • One nickel and one dime => 5 + 10 = 15 cents

The possible values of X are {2, 6, 10, 11, 15}.

To find the probability mass function (pmf), we need to calculate the probabilities of obtaining each sum:

- P(X = 2): Selecting two pennies, which can happen in 1 way out of a total of 10 ways to choose any 2 coins. So, P(X = 2) = 1/10.

- P(X = 6): Selecting one penny and one nickel. There are 2 ways to select this combination, leading to P(X = 6) = 2/10.

- P(X = 10): Selecting two nickels. There is 1 way to choose this combination, so P(X = 10) = 1/10.

- P(X = 11): Selecting one dime and one penny. There are 2 ways to select this combination, so P(X = 11) = 2/10.

- P(X = 15): Selecting one dime and one nickel. There is 1 way to select this, so P(X = 15) = 1/10.

The pmf can be summarized as follows:

  • P(X = 2) = 1/10

  • P(X = 6) = 2/10

  • P(X = 10) = 1/10

  • P(X = 11) = 2/10

  • P(X = 15) = 1/10

Problem 2:

The cumulative distribution function (CDF) can be derived from the pmf:

  • F(2) = P(X ≤ 2) = 1/10

  • F(6) = P(X ≤ 6) = 3/10

  • F(10) = P(X ≤ 10) = 4/10

  • F(11) = P(X ≤ 11) = 6/10

  • F(15) = P(X ≤ 15) = 1

Next, we can compute E(X), the expectance:

E(X) = Σ [x P(X = x)] = (2(1/10) + 6(2/10) + 10(1/10) + 11(2/10) + 15(1/10)). Simplifying gives: E(X) = 7.6 cents.

Problem 3:

The total number of ways to choose 3 applicants from 11 (6 men + 5 women) is C(11, 3) = 165. To compute the pmf of X, which is the number of women in the chosen group:

- P(0 women): Choose 3 men. C(6, 3) = 20. P(X = 0) = 20/165.

- P(1 woman): C(5, 1) C(6, 2) = 5 15 = 75. P(X = 1) = 75/165.

- P(2 women): C(5, 2) C(6, 1) = 10 6 = 60. P(X = 2) = 60/165.

- P(3 women): C(5, 3) = 1. P(X = 3) = 1/165.

The pmf is:

  • P(X = 0) = 20/165

  • P(X = 1) = 75/165

  • P(X = 2) = 60/165

  • P(X = 3) = 10/165

To find the expected number of women:

E(X) = Σ [x P(X = x)] = (0(20/165) + 1(75/165) + 2(60/165) + 3*(1/165)). This simplifies to E(X) = 1.22 women expected.

Problem 4:

In this problem, we begin with determining the value of p by using the property that probabilities must sum to 1:

p + 0.1 + 0.3 + 0.1 + 4p = 1 leads to 6p = 0.5 so p = 0.0833.

For part (b), we compute P(1.9 ≤ |X| ≤ 3): This involves calculating P(X = 1.9) + P(X = 0.1) + P(X = 3).

The cumulative distribution again can be calculated ensuring probabilities sum correctly:

F(0) = 0.1; F(2) = 0.1 + 0.3 + 0.1 = 0.5. Finally, for E(X) where:

E(X) = Σ [x * P(X = x)] can be simplified using the pmf aforementioned. Each computation leads to final expectations needed for further analysis.

References

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