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 x of the time. It is agreed that his claim will be accepted if he correctly predicts the results of at least m of n matches. What is the probability that his claim gets rejected?
Options: a. ∑i=0m−1nCipi(1−p)n−i b. ∑i=0mnCipi(1−p)n−i c. ∑i=1m−1nCipi(1−p)n−i d. ∑i=1mnCipi(1−p)n−i
Answer: a
Solution: P(Correct match prediction)=p=100x Let X∼Binomial(n,p). Probability claim is rejected:
P(X<m)=i=0∑m−1nCipi(1−p)n−iExample: If x=75, m=5, n=6:
P(X<5)=i=0∑46Ci(0.75)i(0.25)6−i≈0.472. Which Statement Is Always True for Binomial Distribution?
Question: If X∼Binomial(n,p), then which of the following statement/s is/are always true? (n>0 and 0<p<1)
Options: a. E(X)≤Var(X) b. E(X)<Var(X) c. E(X)≥Var(X) d. E(X)>Var(X) e. Var(X)≤SD(X) f. Var(X)≥SD(X)
Answer: d
Solution: E(X)=np, Var(X)=np(1−p) Since 0<p<1, np>np(1−p) always, so E(X)>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 x rounds out of the y 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:b. Find the probability that the boxing game will be played on Xbox? (Enter the answer correct to 2 decimal places)
Answer:
i=x∑yyCipi(1−p)y−iwhere p=a+ba
Solution: Let X∼Binomial(y,p), where p=a+ba. Probability boxing game is played:
P(X≥x)=i=x∑yyCipi(1−p)y−iExample: If x=3, y=5, a=3, b=2:
P(X≥3)=i=3∑55Ci(0.6)i(0.4)5−i≈0.684. Finding the Parameter p of Binomial Distribution
Question: Let X∼Binomial(n,p). If the probabilities of x and x+1 successes are a and b respectively, then find the parameter ‘p’ of the distribution. (Enter the answer correct to 2 decimal places)
Answer:
p=a(n−x)+b(x+1)b(x+1)Solution:
ba=nCx+1px+1(1−p)n−x−1nCxpx(1−p)n−xSolve for p:
p=a(n−x)+b(x+1)b(x+1)Example: If n=5, x=1, a=0.4096, b=0.2048:
p=0.4096×4+0.2048×20.2048×2=0.25. Probability of Hitting at Least One Six
Question: If the expected number of sixes hit by a batsman on n balls is e and the variance for the same is v, then what is the probability of him hitting at least one six on any randomly selected n balls? (Enter the answer correct to 4 decimal places)
Answer:
1−(1−p)nSolution: Let X∼Binomial(n,p). Given E(X)=np=e, Var(X)=np(1−p)=v.
p=1−evP(X≥1)=1−(1−p)nExample: If e=4, v=34:
p=1−44/3=32n=pe=6P(X≥1)=1−(31)6≈0.99866. Probability of Clearing Exam at Least Twice
Question: The probability of a student clearing a competitive exam is m1. If he gives the exam n 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(m1)0(1−m1)n+nC1(m1)1(1−m1)n−1]Solution: Let X∼Binomial(n,m1). Probability of clearing at least twice:
P(X≥2)=1−[P(X=0)+P(X=1)]Example: If m=4, n=7:
P(X≥2)=1−[(43)7+7×41×(43)6]≈0.567. Correct Conditions About Binomial Distribution
Question: Choose the correct condition/s about binomial distribution.
Options: a. The probability of success p keeps varying for each trial. b. The number of trials n 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 n 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 X 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=i | P(X=i) |
---|---|
0 | 5C0p0(1−p)5 |
1 | 5C1p1(1−p)4 |
2 | 5C2p2(1−p)3 |
3 | 5C3p3(1−p)2 |
4 | 5C4p4(1−p)1 |
5 | 5C5p5(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: p
Solution: Given P(X=5)=0.0009765625:
p5=0.0009765625⟹p=0.25Probability of head: p=0.25
9. Probability Exactly Three Students Are Punished
Question: At a school function, it is noticed that a of the students are not wearing polished shoes and b 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(10000100×(a+b)−(a×b))3(1000010000−100×(a+b)+(a×b))2 b. 5C3(10000a×b)3(1000010000−a×b)2 c. 5C3(100a+b)3(100100−(a+b))2 d. 5C3(100a)3(100100−a)2+5C3(100b)3(100100−b)2
Answer: a
Solution: Let A: not wearing polished shoes, B: not wearing school tie.
P(A∪B)=100a+100b−10000a×bP(X=3)=5C3(100a+b−10000a×b)3(1−(100a+b−10000a×b))2Example: If a=5, b=10:
P(A∪B)=0.15−0.005=0.145P(X=3)=5C3(0.145)3(0.855)2≈0.022310. Expected Number of Black Pens Chosen
Question: There are x black and y blue pens in a box. A pen is chosen at random, and its color is noted. If the process repeats independently, n times with replacement, then calculate the expected number of black pens chosen.
Options: a. n×(yx) b. n×(x+yx) c. n×(1−yx) d. n×(x+yy)
Answer: b
Solution:
E(X)=n×x+yxExample: If x=10, y=20, n=10:
E(X)=10×3010=31011. 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: X∼Poisson(2)
P(X=0)=e−2≈0.1412. 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 X denotes the number of boys in the quiz team, then using the given information, answer questions (12) and (13).
Calculate P(X=2). (Enter the answer correct to 1 decimal place)
Answer: 0.5 (Range: 0.4 to 0.6)
Solution:
P(X=2)=10C36C2×4C1=12015×4=0.513. Expected Number of Boys in Quiz Team
Question: Calculate the value of E(X). (Enter the answer correct to 1 decimal place)
Answer: 1.8 (Range: 1.5 to 2.1)
Solution:
E(X)=Nmn=106×3=1.8This covers all questions and solutions from the PDF123.