Week 11 Graded Assignment

Week 11 Graded Assignment


Statistics Graded Assignment


1. Probability That a Match Predictor’s Claim Is Rejected

Question: A match predictor claims that he can predict the result of a match correctly xx% of the time. It is agreed that his claim will be accepted if he correctly predicts the results of at least mm of nn matches. What is the probability that his claim gets rejected?

Options: a. i=0m1nCipi(1p)ni\sum_{i=0}^{m-1}{ }^{n}C_{i} p^{i}(1-p)^{n-i} b. i=0mnCipi(1p)ni\sum_{i=0}^{m}{ }^{n}C_{i} p^{i}(1-p)^{n-i} c. i=1m1nCipi(1p)ni\sum_{i=1}^{m-1}{ }^{n}C_{i} p^{i}(1-p)^{n-i} d. i=1mnCipi(1p)ni\sum_{i=1}^{m}{ }^{n}C_{i} p^{i}(1-p)^{n-i}

Answer: a

Solution: P(Correct match prediction)=p=x100P(\text{Correct match prediction}) = p = \frac{x}{100} Let XBinomial(n,p)X \sim \text{Binomial}(n, p). Probability claim is rejected:

P(X<m)=i=0m1nCipi(1p)ni P(X < m) = \sum_{i=0}^{m-1}{ }^{n}C_{i} p^{i}(1-p)^{n-i}

Example: If x=75x = 75, m=5m = 5, n=6n = 6:

P(X<5)=i=046Ci(0.75)i(0.25)6i0.47 P(X < 5) = \sum_{i=0}^{4}{ }^{6}C_{i} (0.75)^{i}(0.25)^{6-i} \approx 0.47

2. Which Statement Is Always True for Binomial Distribution?

Question: If XBinomial(n,p)X \sim \text{Binomial}(n, p), then which of the following statement/s is/are always true? (n>0n > 0 and 0<p<10 < p < 1)

Options: a. E(X)Var(X)E(X) \leq \text{Var}(X) b. E(X)<Var(X)E(X) < \text{Var}(X) c. E(X)Var(X)E(X) \geq \text{Var}(X) d. E(X)>Var(X)E(X) > \text{Var}(X) e. Var(X)SD(X)\text{Var}(X) \leq \text{SD}(X) f. Var(X)SD(X)\text{Var}(X) \geq \text{SD}(X)

Answer: d

Solution: E(X)=npE(X) = np, Var(X)=np(1p)\text{Var}(X) = np(1-p) Since 0<p<10 < p < 1, np>np(1p)np > np(1-p) always, so E(X)>Var(X)E(X) > \text{Var}(X). Options (e) and (f) are not always true.


3. Probability of Boxing Game Being Played

Question: Two friends (say ‘A’ and ‘B’) could not decide whether to play a racing game or a boxing game on Xbox. They decide to play a card game first. If ‘A’ wins at least xx rounds out of the yy rounds of the card game played, then the boxing game will be played. The chances of ‘A’ winning in any round of the card game is a:ba : b. Find the probability that the boxing game will be played on Xbox? (Enter the answer correct to 2 decimal places)

Answer:

i=xyyCipi(1p)yi \sum_{i=x}^{y}{ }^{y}C_{i} p^{i}(1-p)^{y-i}

where p=aa+bp = \frac{a}{a+b}

Solution: Let XBinomial(y,p)X \sim \text{Binomial}(y, p), where p=aa+bp = \frac{a}{a+b}. Probability boxing game is played:

P(Xx)=i=xyyCipi(1p)yi P(X \geq x) = \sum_{i=x}^{y}{ }^{y}C_{i} p^{i}(1-p)^{y-i}

Example: If x=3x = 3, y=5y = 5, a=3a = 3, b=2b = 2:

P(X3)=i=355Ci(0.6)i(0.4)5i0.68 P(X \geq 3) = \sum_{i=3}^{5}{ }^{5}C_{i} (0.6)^{i}(0.4)^{5-i} \approx 0.68

4. Finding the Parameter pp of Binomial Distribution

Question: Let XBinomial(n,p)X \sim \text{Binomial}(n, p). If the probabilities of xx and x+1x+1 successes are aa and bb respectively, then find the parameter ‘p’ of the distribution. (Enter the answer correct to 2 decimal places)

Answer:

p=b(x+1)a(nx)+b(x+1) p = \frac{b(x+1)}{a(n-x) + b(x+1)}

Solution:

ab=nCxpx(1p)nxnCx+1px+1(1p)nx1 \frac{a}{b} = \frac{{ }^{n}C_{x} p^{x}(1-p)^{n-x}}{{ }^{n}C_{x+1} p^{x+1}(1-p)^{n-x-1}}

Solve for pp:

p=b(x+1)a(nx)+b(x+1) p = \frac{b(x+1)}{a(n-x) + b(x+1)}

Example: If n=5n = 5, x=1x = 1, a=0.4096a = 0.4096, b=0.2048b = 0.2048:

p=0.2048×20.4096×4+0.2048×2=0.2 p = \frac{0.2048 \times 2}{0.4096 \times 4 + 0.2048 \times 2} = 0.2

5. Probability of Hitting at Least One Six

Question: If the expected number of sixes hit by a batsman on nn balls is ee and the variance for the same is vv, then what is the probability of him hitting at least one six on any randomly selected nn balls? (Enter the answer correct to 4 decimal places)

Answer:

1(1p)n 1 - (1-p)^{n}

Solution: Let XBinomial(n,p)X \sim \text{Binomial}(n, p). Given E(X)=np=eE(X) = np = e, Var(X)=np(1p)=v\text{Var}(X) = np(1-p) = v.

p=1ve p = 1 - \frac{v}{e} P(X1)=1(1p)n P(X \geq 1) = 1 - (1-p)^{n}

Example: If e=4e = 4, v=43v = \frac{4}{3}:

p=14/34=23 p = 1 - \frac{4/3}{4} = \frac{2}{3} n=ep=6 n = \frac{e}{p} = 6 P(X1)=1(13)60.9986 P(X \geq 1) = 1 - \left(\frac{1}{3}\right)^6 \approx 0.9986

6. Probability of Clearing Exam at Least Twice

Question: The probability of a student clearing a competitive exam is 1m\frac{1}{m}. If he gives the exam nn times, then what is the probability of him clearing the exam at least twice? (Enter the answer correct to 2 decimal places)

Answer:

1[nC0(1m)0(11m)n+nC1(1m)1(11m)n1] 1 - \left[{ }^{n}C_{0}\left(\frac{1}{m}\right)^{0}\left(1-\frac{1}{m}\right)^{n} + { }^{n}C_{1}\left(\frac{1}{m}\right)^{1}\left(1-\frac{1}{m}\right)^{n-1}\right]

Solution: Let XBinomial(n,1m)X \sim \text{Binomial}(n, \frac{1}{m}). Probability of clearing at least twice:

P(X2)=1[P(X=0)+P(X=1)] P(X \geq 2) = 1 - [P(X=0) + P(X=1)]

Example: If m=4m = 4, n=7n = 7:

P(X2)=1[(34)7+7×14×(34)6]0.56 P(X \geq 2) = 1 - \left[\left(\frac{3}{4}\right)^7 + 7 \times \frac{1}{4} \times \left(\frac{3}{4}\right)^6\right] \approx 0.56

7. Correct Conditions About Binomial Distribution

Question: Choose the correct condition/s about binomial distribution.

Options: a. The probability of success pp keeps varying for each trial. b. The number of trials nn is finite. c. The trials are dependent on each other. d. The trials are independent of each other.

Answer: b, d

Solution:

  • Number of trials nn is fixed.
  • Trials are independent.

8. Probability of Getting a Head (Coin Toss)

Question: Rithika wants to test whether the coin she has is a fair coin or not. To test this, she conducted an experiment of tossing the coin 5 times. Binomial random variable XX is defined as the total number of heads after 5 tosses. The probability distribution of the binomial random variable is given in Table 11.1.G.

X=iX = iP(X=i)P(X = i)
05C0p0(1p)5{ }^{5}C_{0} p^{0}(1-p)^{5}
15C1p1(1p)4{ }^{5}C_{1} p^{1}(1-p)^{4}
25C2p2(1p)3{ }^{5}C_{2} p^{2}(1-p)^{3}
35C3p3(1p)2{ }^{5}C_{3} p^{3}(1-p)^{2}
45C4p4(1p)1{ }^{5}C_{4} p^{4}(1-p)^{1}
55C5p5(1p)0{ }^{5}C_{5} p^{5}(1-p)^{0}

What is the approximate probability of getting a head in tossing the given coin? (Enter the answer correct to one decimal place)

Answer: pp

Solution: Given P(X=5)=0.0009765625P(X=5) = 0.0009765625:

p5=0.0009765625    p=0.25 p^5 = 0.0009765625 \implies p = 0.25

Probability of head: p=0.25p = 0.25


9. Probability Exactly Three Students Are Punished

Question: At a school function, it is noticed that aa% of the students are not wearing polished shoes and bb% of students are not wearing school ties. It is announced that the students who have committed any of the infractions will be punished, and that these two infractions are independent of one another. If a teacher selects 5 students at random, then find the probability that exactly three of the students will be punished for any of the infractions?

Options: a. 5C3(100×(a+b)(a×b)10000)3(10000100×(a+b)+(a×b)10000)2{ }^{5}C_{3}\left(\frac{100 \times (a+b) - (a \times b)}{10000}\right)^{3}\left(\frac{10000 - 100 \times (a+b) + (a \times b)}{10000}\right)^{2} b. 5C3(a×b10000)3(10000a×b10000)2{ }^{5}C_{3}\left(\frac{a \times b}{10000}\right)^{3}\left(\frac{10000 - a \times b}{10000}\right)^{2} c. 5C3(a+b100)3(100(a+b)100)2{ }^{5}C_{3}\left(\frac{a+b}{100}\right)^{3}\left(\frac{100 - (a+b)}{100}\right)^{2} d. 5C3(a100)3(100a100)2+5C3(b100)3(100b100)2{ }^{5}C_{3}\left(\frac{a}{100}\right)^{3}\left(\frac{100 - a}{100}\right)^{2} + { }^{5}C_{3}\left(\frac{b}{100}\right)^{3}\left(\frac{100 - b}{100}\right)^{2}

Answer: a

Solution: Let AA: not wearing polished shoes, BB: not wearing school tie.

P(AB)=a100+b100a×b10000 P(A \cup B) = \frac{a}{100} + \frac{b}{100} - \frac{a \times b}{10000} P(X=3)=5C3(a+b100a×b10000)3(1(a+b100a×b10000))2 P(X=3) = { }^{5}C_{3}\left(\frac{a+b}{100} - \frac{a \times b}{10000}\right)^{3}\left(1 - \left(\frac{a+b}{100} - \frac{a \times b}{10000}\right)\right)^{2}

Example: If a=5a = 5, b=10b = 10:

P(AB)=0.150.005=0.145 P(A \cup B) = 0.15 - 0.005 = 0.145 P(X=3)=5C3(0.145)3(0.855)20.0223 P(X=3) = { }^{5}C_{3} (0.145)^{3} (0.855)^{2} \approx 0.0223

10. Expected Number of Black Pens Chosen

Question: There are xx black and yy blue pens in a box. A pen is chosen at random, and its color is noted. If the process repeats independently, nn times with replacement, then calculate the expected number of black pens chosen.

Options: a. n×(xy)n \times \left(\frac{x}{y}\right) b. n×(xx+y)n \times \left(\frac{x}{x+y}\right) c. n×(1xy)n \times \left(1 - \frac{x}{y}\right) d. n×(yx+y)n \times \left(\frac{y}{x+y}\right)

Answer: b

Solution:

E(X)=n×xx+y E(X) = n \times \frac{x}{x+y}

Example: If x=10x = 10, y=20y = 20, n=10n = 10:

E(X)=10×1030=103 E(X) = 10 \times \frac{10}{30} = \frac{10}{3}

11. Probability No Call Is Received (Poisson)

Question: Number of calls received at an office follows a Poisson distribution with an average of 1 call per minute. Find the probability that no call will be received in two minutes at the office. (Enter the answer correct to 2 decimal accuracy.)

Answer: 0.14 (Range: 0.10 to 0.16)

Solution: XPoisson(2)X \sim \text{Poisson}(2)

P(X=0)=e20.14 P(X=0) = e^{-2} \approx 0.14

12. Probability of Exactly Two Boys in Quiz Team

Question: A quiz team is to be chosen randomly from 6 boys and 4 girls. The team has 3 slots which are to be filled randomly. If XX denotes the number of boys in the quiz team, then using the given information, answer questions (12) and (13).

Calculate P(X=2)P(X=2). (Enter the answer correct to 1 decimal place)

Answer: 0.5 (Range: 0.4 to 0.6)

Solution:

P(X=2)=6C2×4C110C3=15×4120=0.5 P(X=2) = \frac{{ }^{6}C_{2} \times { }^{4}C_{1}}{{ }^{10}C_{3}} = \frac{15 \times 4}{120} = 0.5

13. Expected Number of Boys in Quiz Team

Question: Calculate the value of E(X)E(X). (Enter the answer correct to 1 decimal place)

Answer: 1.8 (Range: 1.5 to 2.1)

Solution:

E(X)=mnN=6×310=1.8 E(X) = \frac{m n}{N} = \frac{6 \times 3}{10} = 1.8

This covers all questions and solutions from the PDF123.