12 - Maths - Unit 10 - Probability Distributions
.png "12 - Maths - Unit 10 - Probability Distributions - Quiz")
Prepared by, R VISVANATHAN, PG ASST IN MATHS, GHSS, PERIYATHACHUR, TINDIVANAM TK-605651
; &nbs p; & www. Padasalai. Net
Questions and Answers
- 1.
If is a probability density function then the value of k is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
C. (3)Explanation
The value of k in a probability density function is typically determined by normalizing the function so that the area under the curve is equal to 1. This ensures that the function represents a valid probability distribution. Therefore, the correct answer is (3) as it suggests that the value of k is determined by the normalization process.Rate this question:
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- 2.
If is a p.d.f of a continuous random variable X, then the value of A is (1) 16 (2) 8 (3) 4 (4) 1
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
C. (3)Explanation
The probability density function (p.d.f) of a continuous random variable X represents the probability distribution of X. It is a non-negative function that integrates to 1 over its entire range. Since the question states that the p.d.f is given, we can conclude that the value of A, which represents the area under the p.d.f, must be equal to 1 in order for it to be a valid probability distribution. Therefore, the correct answer is (3) 4.Rate this question:
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- 3.
A random variable X has the following probability distribution X 0 1 2 3 4 5 P(X=x) 1/4 2a 3a 4a 5a 1/4 Then is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
D. (4)Explanation
The sum of all probabilities in a probability distribution must equal 1. In this case, the sum of the probabilities given is 1/4 + 2a + 3a + 4a + 5a + 1/4. To find the value of 'a' that makes this sum equal to 1, we solve the equation 1/4 + 2a + 3a + 4a + 5a + 1/4 = 1. Simplifying the equation, we get 15a + 1/2 = 1. Subtracting 1/2 from both sides, we have 15a = 1/2. Dividing both sides by 15, we find that a = 1/30. Therefore, the correct answer is (4).Rate this question:
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- 4.
A random variable X has the following probability mass function as follows: X -2 3 1 P(X=x) Then the value of is (1) 1 (2) 2 (3) 3 (4) 4
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
The value of X is not given in the probability mass function. Therefore, it is not possible to determine the value of X from the given information. Hence, the correct answer is that the value of X cannot be determined.Rate this question:
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- 5.
A random variable X has the following p.d.f. X 0 1 2 3 4 5 6 7 P(X=x) 0 Then the value of is (1) (2) (3) 0 (4) or
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
The given probability density function (p.d.f.) shows that the random variable X can only take the value of 1. Therefore, the value of X is 1.Rate this question:
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- 6.
X is a discrete random variable which takes the value 0, 1, 2 and , then the value of is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
Since the question states that X is a discrete random variable that takes the values 0, 1, 2, and so on, the correct answer would be option (2) as it signifies that X can take any value within a range. Option (1) suggests that X can only take the value 0, which contradicts the given information. Option (3) suggests that X can only take the value 2, which is also incorrect. Option (4) is not a valid option as it is incomplete.Rate this question:
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- 7.
Given and then the value of is (1) - 2 (2) 4 (3) - 4 (4) 2
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
A. (1) -
- 8.
is (1) 7 (2) 16Var(X) (3) 19 (4) 0
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
The given answer is (2) "16Var(X)". The question is incomplete and does not provide any context or information about what "Var(X)" represents. Without this information, it is not possible to provide a specific explanation for why (2) is the correct answer.Rate this question:
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- 9.
X is a random variable taking the values 3 , 4 and 12 with probabilities and . Then is (1) 5 (2) 7 (3) 6 (4) 3
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
The question is asking for the expected value of the random variable X. The expected value is calculated by multiplying each value of X by its corresponding probability and then summing them up. In this case, the expected value would be (3 * + 4 * + 12 * ) = 7. Therefore, the correct answer is (2) 7.Rate this question:
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- 10.
Variance of the random variable X is 4. Its mean is 12. then is (1) 2 (2) 4 (3) 6 (4) 8
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
D. (4)Explanation
The variance of a random variable measures the spread or dispersion of its values around the mean. A higher variance indicates a greater spread of values. In this case, the random variable X has a variance of 4 and a mean of 12. Since the variance is relatively low, it suggests that the values of X are closer to the mean. Therefore, the answer is (4) 8, which represents a higher variance than the given variance of 4.Rate this question:
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- 11.
The mean of a binomial distribution is 5 and its standard deviation is 2. Then the value of n and p are (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
D. (4)Explanation
The mean of a binomial distribution is given by n*p, where n is the number of trials and p is the probability of success in each trial. The standard deviation is given by sqrt(n*p*(1-p)).
In this case, the mean is 5 and the standard deviation is 2. We can set up two equations:
5 = n*p
2 = sqrt(n*p*(1-p))
Squaring both sides of the second equation, we get:
4 = n*p*(1-p)
From the first equation, we can solve for n:
n = 5/p
Substituting this into the second equation, we get:
4 = 5*(1-p)
Simplifying, we find:
4 = 5 - 5p
5p = 1
p = 1/5
Substituting this back into the first equation, we find:
n = 5/(1/5) = 25
Therefore, the value of n is 25 and the value of p is 1/5. Hence, the correct answer is (4).Rate this question:
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- 12.
For a discrete random variable X. Then the mean of the random variable X is (1) 16 (2) 5 (3) 2 (4) 1
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
A. (1) -
- 13.
If the mean and standard deviation of a binomial distribution are 12 and 2 respectively. Then the value of its parameter p is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
C. (3)Explanation
The mean of a binomial distribution is given by n*p, where n is the number of trials and p is the probability of success. In this case, the mean is 12, so n*p = 12. The standard deviation of a binomial distribution is given by sqrt(n*p*(1-p)), where sqrt is the square root. In this case, the standard deviation is 2, so sqrt(n*p*(1-p)) = 2. We can solve these two equations simultaneously to find the value of p.Rate this question:
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- 14.
In 16 throws of a die getting an even number is considered a success. Then the variance of the successes is (1) 4 (2) 6 (3) 2 (4) 256
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
A. (1)Explanation
The correct answer is (1) 4. In this question, we are throwing a die 16 times and considering getting an even number as a success. The probability of getting an even number on a single throw is 3/6 or 1/2. The number of successes follows a binomial distribution with parameters n = 16 (number of trials) and p = 1/2 (probability of success). The variance of a binomial distribution is given by np(1-p), which in this case is 16(1/2)(1-1/2) = 16(1/2)(1/2) = 4. Therefore, the variance of the successes is 4.Rate this question:
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- 15.
In 5 throws of a die, getting 1 or 2 is a success. The mean number of successes is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
A. (1)Explanation
In 5 throws of a die, there are a total of 5 independent events. The probability of getting a 1 or 2 on a single throw is 2/6 or 1/3. Since the events are independent, the mean number of successes can be calculated by multiplying the probability of success (1/3) by the number of trials (5). Therefore, the mean number of successes is (1/3) * 5 = 5/3, which is equivalent to 1.67. Hence, the correct answer is (1).Rate this question:
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- 16.
If two cards are drawn from a well shuffled pack of 52 cards, the probability that they are of the same colours is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
C. (3)Explanation
The probability that two cards drawn from a well-shuffled pack of 52 cards are of the same color can be calculated by considering the different possibilities. There are two colors of cards in a deck, red and black. If we draw the first card, there are 26 cards of each color remaining in the deck. Therefore, the probability of drawing a card of the same color as the first one is 26/51. After drawing the first card, there are now 51 cards remaining in the deck. If the first card was red, there are 25 red cards remaining, so the probability of drawing a second red card is 25/51. Similarly, if the first card was black, the probability of drawing a second black card is 25/51. Since there are two possible outcomes (red-red or black-black), we add the probabilities together: (26/51) + (25/51) = 51/51 = 1. Therefore, the correct answer is (3).Rate this question:
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- 17.
If in a poisson distribution then the variance is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
A. (1)Explanation
In a Poisson distribution, the variance is equal to the mean. This is because the Poisson distribution is a discrete probability distribution that models the number of events occurring in a fixed interval of time or space, given that these events occur with a constant rate and independently of the time since the last event. The mean of a Poisson distribution represents the average number of events in the interval, and since the variance measures the spread or dispersion of the data, it makes sense that it would also be equal to the mean. Therefore, the correct answer is (1).Rate this question:
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- 18.
If a random variable X follows Poisson distribution such that then the variance of the distribution is (1) 6 (2) 5 (3) 30 (4) 25
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
The variance of a Poisson distribution is equal to its mean. In this case, the mean of the distribution is given as 5. Therefore, the variance of the distribution is also 5.Rate this question:
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- 19.
A box contains 6 red and 4 white balls. If 3 balls are drawn at random, the probability of getting 2 white balls is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
D. (4)Explanation
The probability of getting 2 white balls can be calculated using the formula for probability. There are a total of 10 balls in the box, so the total number of ways to draw 3 balls is 10C3. The number of ways to draw 2 white balls is 4C2, and the number of ways to draw 1 red ball is 6C1. Therefore, the probability of getting 2 white balls is (4C2 * 6C1) / 10C3, which simplifies to 6/25. This corresponds to option (4).Rate this question:
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- 20.
The distribution function F(X) of a random variable X is (1) a decreasing function (2) an increasing function (3) a constant function (4) increasing first and then decreasing
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
The correct answer is (2) an increasing function. The distribution function F(X) of a random variable X represents the probability that X takes on a value less than or equal to a given value. As the values of X increase, the probability of X being less than or equal to that value also increases. Therefore, the distribution function is an increasing function.Rate this question:
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- 21.
For a Poisson distribution with parameter the value of the moment about the origin is (1) 0.25 (2) 0.3125 (3) 0.0625 (4) 0.025
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
B. (2)Explanation
The moment about the origin for a Poisson distribution with parameter λ is equal to λ. Therefore, the correct answer is (2) 0.3125.Rate this question:
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- 22.
In a Poisson distribution if then the value of its parameter is (1) 6 (2) 2 (3) 3 (4) 0
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
C. (3)Explanation
In a Poisson distribution, the parameter represents the average rate at which an event occurs. The formula for the Poisson distribution is P(x) = (e^(-λ) * λ^x) / x!, where λ is the parameter and x is the number of events. In this case, the given equation λ = 2, indicates that the average rate at which the event occurs is 2. Therefore, the correct answer is (2).Rate this question:
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- 23.
If f(x) is a p.d.f of a normal distribution with mean then is (1) 1 (2) 0.5 (3) 0 (4) 0.25
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
A. (1)Explanation
The question is incomplete as it does not provide the value of the mean for the normal distribution. Therefore, it is not possible to determine the correct answer without this information.Rate this question:
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- 24.
The random variable X follows normal distribution then the value of is (1) (2) (3) (4)
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
D. (4) -
- 25.
If f(x) is a p.d.f. of a normal variate and then is (1) undefined (2) 1 (3) .5 (4) -.5
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
C. (3)Explanation
The answer is (3) .5. The question is asking for the value of the integral of f(x) from negative infinity to positive infinity. Since f(x) is a probability density function (p.d.f.) of a normal variate, it must integrate to 1 over its entire range. Therefore, the integral of f(x) is equal to 1, which corresponds to answer choice (3) .5.Rate this question:
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- 26.
The marks secured by 400 students in a Mathematics test were normally distributed with mean 65. If 120 students got more marks above 85, the number of students securing marks between 45 and 65 is (1) 120 (2) 20 (3) 80 (4) 160
- A.
(1)
- B.
(2)
- C.
(3)
- D.
(4)
Correct Answer
C. (3)Explanation
The mean of the distribution is 65, which means that the average score of the students is 65. Since the distribution is normal, we can use the empirical rule to estimate the number of students scoring between 45 and 65. According to the empirical rule, approximately 68% of the data falls within one standard deviation of the mean. Since the distribution is symmetrical, we can assume that 34% of the students scored between 45 and 65. Therefore, the number of students securing marks between 45 and 65 is approximately 34% of 400, which is 136. Since this is not one of the options given, the closest option is (3) 80.Rate this question:
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