Week 9 Graded Assignment

Statistics Graded Assignment

1. Value of k in Cumulative Distribution Function

Question: A discrete random variable $ X $ can take the values $ 1,2,3,···,n $. For these values the cumulative distribution function is defined by:

$$ F(x) = P(X < x) = \frac{x^2 + k}{m}; \quad x = 1, 2, 3, \ldots, n $$

Find the value of $ k $.

Answer:

$$ k = m - n^2 $$

Correction: The solution in the PDF says $ k = m - n^2/2 $, but the calculation uses $ F(n) = 1 $, so $ \frac{n^2 + k}{m} = 1 \implies k = m - n^2 $. However, the PDF solution says:

$$ F(n) = \frac{n^2 + k}{m} = 1 \implies k = m - n^2 $$

But in the numerical example, with $ n=3, m=40 $, $ \frac{3^2 + k}{40} = 1 $ gives $ k = 31 $, which matches $ m - n^2 = 31 $. However, the PDF answer is written as $ k = m - n^2/2 $, but the calculation and example clearly use $ k = m - n^2 $.

Solution:

$$ F(n) = 1 \implies \frac{n^2 + k}{m} = 1 \implies k = m - n^2 $$

Example: For $ n=3, m=40 $:

$$ \frac{9 + k}{40} = 1 \implies k = 31 $$

Note: The PDF answer is inconsistent with the calculation and example, but the correct formula is $ k = m - n^2 $.

2. Probability that Net Gain from Ticket Purchase is Less than b

Question:

An organization in Texas organizes a lucky draw this month. $ n $ thousand tickets are sold for $ m $ each. Each has an equal chance of winning. $ x $ tickets will win $ a$ $, $ y $ tickets will win $ b$ $, and $ z $ tickets will win $ c$ $. Let the random variable $ X $ denote the net gain from purchase of one ticket. What is the probability that $ X $ takes a value less than $ b $?

Answer:

$$ P(X < b) = \frac{n \times 1000 - x}{n \times 1000} $$

Correction: The solution in the PDF is unclear and has typographical errors. The correct interpretation is:

$$ P(X < b) = P(\text{win less than } b) + P(\text{no win}) = \frac{z + (n \times 1000 - x - y - z)}{n \times 1000} $$

But the PDF says:

$$ P(X < b) = \frac{n \times 1000 - x}{n \times 1000} $$

But this is not correct unless $ b $ is the highest prize above which only $ x $ tickets win. The solution likely intends to ask for the probability that the net gain is less than $ b - m $ (i.e., not winning the top prize $ a $), but the question is ambiguous. In the example, $ P(X < 500) = 0.9998 $.

Solution:

$$ P(X < b) = \frac{n \times 1000 - x}{n \times 1000} $$

Example: For $ n=5, x=1, y=2, z=10 $:

$$ P(X < 500) = \frac{2 + 10 + 4987}{5000} = \frac{4999}{5000} = 0.9998 $$

Note: The question and answer are not clearly matched, but the example calculation is correct for the question as interpreted in the PDF.

3. Number of Possible Values for a Random Variable

Question: In a group of $ n $ people, $ x $ are photographers and $ n - x $ are journalists. $ m $ people are randomly picked from a group of these $ n $ people. Let $ Y $ be a random variable which represents the number of photographers. How many possible values can the random variable $ Y $ take?

Answer:

$$ m + 1 $$

Solution: Possible values of $ Y $ are $ 0, 1, 2, ···, m $. Hence, the number of possible values is $ m + 1 $.

Example: For $ m=8, x=240, n=15 $: Possible values: $ 0,1,2,···,8 $, so $ 9 $ values.

4. Discrete Random Variables

Question: Which of the following is/are discrete random variables?

a. Number of tires produced in an automotive tire factory every 30 minutes. b. The number of kernels of popcorn in a 1 kg container. c. The time between customers entering a checkout lane at a retail store. d. The amount of rain recorded at an airport one day. e. The amount of liquid in a 2 litres bottle of soft drink. f. The number of no-shows for every 1000 reservations made with a commercial airline.

Answer: a, b, f

Solution: a, b, and f have countable possible values, so they are discrete. c, d, and e can take any value in an interval, so they are continuous.

5. Probability Mass Function for Head Runs

Question: A biased coin with probability of heads 0.75 is tossed three times. Let $ X $ be a random variable that represents the number of head runs, a head run being defined as a consecutive occurrence of at least two heads. Then the probability mass function of $ X $ is given by:

$$ P(X = x) = \begin{cases} 0.375 & \text{for } x = 0 \\ 0.625 & \text{for } x = 1 \end{cases} $$

$$ P(X = x) = \begin{cases} 0.297 & \text{for } x = 0 \\ 0.703 & \text{for } x = 1 \end{cases} $$

$$ P(X = x) = \begin{cases} 0.016 & \text{for } x = 0 \\ 0.140 & \text{for } x = 1 \\ 0.422 & \text{for } x = 2 \\ 0.422 & \text{for } x = 3 \end{cases} $$

$$ P(X = x) = \begin{cases} 0.016 & \text{for } x = 0 \\ 0.844 & \text{for } x = 1 \\ 0.140 & \text{for } x = 2 \end{cases} $$

Answer: b

Solution:

HHH: 1 head run, probability $ 0.422 $
HHT: 1 head run, probability $ 0.141 $
HTH: 0 head runs, probability $ 0.141 $
HTT: 0 head runs, probability $ 0.047 $
THH: 1 head run, probability $ 0.141 $
THT: 0 head runs, probability $ 0.047 $
TTH: 0 head runs, probability $ 0.047 $
TTT: 0 head runs, probability $ 0.016 $

$$ P(X = 0) = 0.141 + 0.047 + 0.047 + 0.047 + 0.016 = 0.297 $$$$ P(X = 1) = 0.422 + 0.141 + 0.141 = 0.703 $$

Hence, option b is correct.

6. Probability that X Takes Value at Most n-1

Question: Nina has $ n $ music sessions in a week. She attends the sessions $ n $ days a week $ x% $ of the time, $ n-1 $ days $ y% $ of the time, one day $ z% $ of the time, and no days $ p% $ of the time. Let $ X $ be a discrete random variable representing the number of sessions she attends in a week. Suppose one week is randomly selected, what is the probability that the random variable $ X $ takes the value at most $ n-1 $?

Answer:

$$ 1 - \frac{x}{100} $$

Solution:

$$ P(X \leq n-1) = P(X = 0) + P(X = 1) + \ldots + P(X = n-1) = 1 - P(X = n) $$$$ P(X = n) = \frac{x}{100} $$$$ P(X \leq n-1) = 1 - \frac{x}{100} $$

7. Value of k for PMF

Question: Find the value of $ k $ for which $ k \left(\frac{m}{n}\right)^x $ ($ x = 0, 1, 2, ··· $) is a probability mass function (pmf).

Answer:

$$ k = \frac{n - m}{n} $$

Solution:

$$ \sum_{x=0}^{\infty} k \left(\frac{m}{n}\right)^x = 1 \implies k \frac{1}{1 - \frac{m}{n}} = 1 \implies k = 1 - \frac{m}{n} $$

But in the PDF:

$$ k = \frac{n - m}{n} $$

This is correct.

Example: For $ m=3, n=8 $:

$$ k = \frac{5}{8} $$

8. Probability $ P(X = 2) $

Question: Using the information in the previous question, calculate $ P(X = 2) $.

Answer:

$$ P(X = 2) = \frac{n - m}{n} \left(\frac{m}{n}\right)^2 $$

Solution:

$$ P(X = 2) = k \left(\frac{m}{n}\right)^2 = \frac{n - m}{n} \left(\frac{m}{n}\right)^2 $$

9. Value of Random Variable with Least Probability

Question: From a box A containing 3 white and 6 black balls, 5 balls are transferred into an empty box B. Let $ X $ be a random variable that represents the number of white balls which are transferred from A to B. What value of random variable will have the least probability?

Answer: 0

Solution:

$$ P(X = 0) = \frac{6C5}{9C5} = 0.048 $$$$ P(X = 1) = \frac{3C1 \times 6C4}{9C5} = 0.357 $$$$ P(X = 2) = \frac{3C2 \times 6C3}{9C5} = 0.476 $$$$ P(X = 3) = \frac{3C3 \times 6C2}{9C5} = 0.119 $$

Thus, $ X = 0 $ has the least probability.

10. Value of k in PMF

Question: The probability mass function of a random variable $ X $ is given by:

$$ P(X = x) = \begin{cases} 3k^2 - 3k & \text{for } x = 0 \\ 2k^2 - 1 & \text{for } x = 1 \\ 0 & \text{otherwise} \end{cases} $$

Determine the value of $ k $ given $ k > 0 $.

Answer: 1

Solution:

$$ (3k^2 - 3k) + (2k^2 - 1) = 1 \implies 5k^2 - 3k - 2 = 0 $$$$ (5k + 2)(k - 1) = 0 \implies k = 1 \text{ (since } k > 0) $$

This covers all questions and solutions from the PDF¹. Note: For Question 1, the correct formula should be $ k = m - n^2 $, not $ k = m - n^2/2 $, as per the calculation and example in the PDF.

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