F. Y. B. Sc. IT Sem - II Internal Examination 2019 - Nsm

In binomial distribution, p represents the probability of success. This means that p is the likelihood of achieving a desired outcome or event in a given number of trials or experiments. It is used to calculate the probability of obtaining a specific number of successes in a fixed number of independent trials, where each trial has only two possible outcomes - success or failure. Therefore, p is the probability of achieving the desired outcome or success in the binomial distribution.

The given function P(x) = 0.1, 0.2, 0.3, 0.2, 0.2 can be considered a probability mass function (p.m.f) because it satisfies the properties of a p.m.f. A p.m.f should have non-negative probabilities for all values of x, and the sum of all probabilities should equal 1. In this case, all the probabilities are non-negative and the sum of probabilities (0.1 + 0.2 + 0.3 + 0.2 + 0.2) equals 1. Therefore, it can be classified as a p.m.f.

The positive square root of variance is called the standard deviation. It is a measure of how spread out the data points in a dataset are. By taking the square root of variance, we can obtain a value that is in the same unit as the original data, making it easier to interpret. Standard deviation is widely used in statistics and is often used to describe the variability or dispersion of a dataset.

A random variable x is said to follow a discrete uniform distribution if its probability mass function (p.m.f.) is equal to 1 divided by the number of possible outcomes (n). This means that each outcome has an equal probability of occurring.

The expected value, E(x), of a binomial distribution is calculated by multiplying the number of trials, n, by the probability of success, p. In this case, n is given as 20 and p is given as 1/10. Multiplying these values together gives us 20 * 1/10 = 2. Therefore, the correct answer is 2.

In Binomial distribution, q is called the probability of failure. This means that q represents the likelihood of an event not occurring or being unsuccessful. In the context of the Binomial distribution, q is used to calculate the probability of a certain number of failures in a fixed number of independent trials, where the probability of success is represented by p. Therefore, q complements the probability of success and helps in determining the probability distribution of failures in a binomial experiment.

The correct answer for this question is npq. In binomial distribution, Var(x) represents the variance of the random variable x. The formula to calculate the variance in a binomial distribution is npq, where n is the number of trials, p is the probability of success in each trial, and q is the probability of failure in each trial (q = 1 - p). This formula allows us to measure the spread or dispersion of the binomial distribution.

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The given function represents a probability mass function (p.m.f) where the values of p(x) are proportional to k, 2k, 3k, 4k, and 5k. In a p.m.f, the sum of all probabilities must equal 1. Since there are 5 possible outcomes in this function, the sum of the probabilities is 1/15 + 2/15 + 3/15 + 4/15 + 5/15 = 15/15 = 1. Therefore, k must be such that the sum of the probabilities equals 1, which is satisfied when k = 1/15.

To find p(2

In a uniform distribution, the variance of a random variable x is equal to (n^2 - 1)/12, where n is the number of possible outcomes. This formula calculates the spread or dispersion of the data points around the mean. The (n^2 - 1)/12 formula is derived from the properties of a uniform distribution and is commonly used to determine the variability in a dataset that follows a uniform distribution.

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The correct answer is 1/λ2. In the exponential distribution, the variance of a random variable x is equal to the reciprocal of the square of the rate parameter (λ). Therefore, the variance is 1/λ2.

The standard deviation is a measure of the dispersion or spread of a set of data. It is calculated as the square root of the variance. In this question, the variance of x is given as 1/6. To find the standard deviation, we take the square root of the variance. Therefore, the standard deviation is √(1/6) which is approximately 0.4082.

A probability mass function (p.m.f.) is a function that assigns probabilities to each possible value of a discrete random variable. In this case, the given function P(x) assigns probabilities of 0.5, -0.1, 0.6, and 0 to the possible values of x. Since all the probabilities are non-negative and their sum is equal to 1 (0.5 + (-0.1) + 0.6 + 0 = 1), it satisfies the properties of a p.m.f. Therefore, the given function is indeed a p.m.f.

To find p(x>=3), we need to add the probabilities of all values of x that are greater than or equal to 3. From the given information, we know that p(3) = 0.15 and p(4) = 0.25. Adding these probabilities gives us 0.15 + 0.25 = 0.4. Therefore, the probability of x being greater than or equal to 3 is 0.4.

To find p(1

The probability function p(x) is given for x=0,1,2,3,4. We can see that p(0) = c, p(1) = 2c, p(2) = 4c, p(3) = 2c, and p(4) = c. To find p(x

The correct answer is 1/λ because in exponential distribution, the expected value (E(x)) is equal to the reciprocal of the rate parameter (λ). This means that on average, the time between events occurring in an exponential distribution is equal to 1/λ.

The probability of getting a head when tossing a fair coin is 1/2. This is because there are two equally likely outcomes when tossing a coin - either it lands on heads or tails. Since the coin is fair, each outcome has an equal chance of occurring, so the probability of getting a head is 1 out of 2, or 1/2.

The given formula calculates the value of q, which represents the probability of the complement of the event occurring in a binomial distribution. By subtracting p from 1, we can find the probability of the event not happening (complement) and therefore calculate q.

In a uniform distribution, all values have equal probability of occurring. The expected value (E(x)) represents the average value of the distribution. In this case, the expected value can be calculated by taking the sum of all possible values and dividing it by the total number of values. Since the values in a uniform distribution range from 1 to n, the sum of all values can be calculated using the formula (n * (n+1))/2. Dividing this sum by the total number of values (n) gives us (n+1)/2, which is the correct answer.

The expected value of a binomial distribution, denoted as E(x), is equal to the product of the number of trials (n) and the probability of success in each trial (p). Therefore, the correct answer is np.

The probability distribution is given by p(0) = c, p(1) = 2c, p(2) = 4c, p(3) = 2c, p(4) = c. To find p(x>=2), we need to sum up the probabilities of all values greater than or equal to 2. In this case, p(2) + p(3) + p(4) = 4c + 2c + c = 7c. Since the total probability must sum up to 1, we have 7c = 1. Solving for c, we get c = 1/7. Therefore, p(x>=2) = 7c = 7/7 = 7/10.

The correct answer for the Poison distribution is m. This is because E(x) represents the expected value or the average number of events in a given interval, and in the Poison distribution, this average is equal to the parameter m.

The given information shows that the function p(x) has a pattern where the values repeat in a symmetric manner. The values of p(x) at x=0, 3, and 4 are all equal to c, while the values at x=1 and 2 are equal to 2c and 4c respectively. This pattern suggests that the function p(x) is a periodic function with a period of 4. The values of p(x) at x=0, 1, 2, 3, and 4 form a cycle that repeats. Since the values of p(x) at x=0, 1, 2, 3, and 4 are c, 2c, 4c, 2c, and c respectively, we can see that c = 1/10.

The mean of an exponential distribution with parameter λ is equal to 1/λ. In this case, the parameter is 0.5, so the mean is 1/0.5 = 2.

The expected value, denoted as E(x), in a rectangular distribution is calculated by taking the average of the lower limit (a) and the upper limit (b) of the distribution. Therefore, the correct answer is (b+a)/2.

The cumulative distribution function (c.d.f.) gives the probability that a random variable takes on a value less than or equal to a particular value. In this case, we are finding the c.d.f. of 1. The given probabilities are p(1) = 3/5, p(3) = 3/10, and p(5) = 1/10. Since we are looking for the probability of a value less than or equal to 1, we only need to consider p(1) which is 3/5. Therefore, the c.d.f. of 1 is 3/5.

In Poison distribution, the variance (Var(x)) is equal to the mean (m). This means that the spread or variability of the data is equal to the average value. In other words, the variance is not affected by the values of p and q, which represent the probability of success and failure respectively. Therefore, the correct answer is m.

Expected value is also called as mean. The mean represents the average value of a set of numbers. In statistics, it is calculated by summing up all the values in a dataset and dividing it by the total number of values. The expected value is used to estimate the long-term average outcome of a random variable or a probability distribution. It is a central concept in probability theory and is often used to make predictions or analyze data.

A continuous random variable x is said to follow a rectangular distribution over an interval (a, b) if it has a probability density function (p.d.f.) equal to 1 divided by the difference between b and a. This means that the probability of any value within the interval (a, b) is constant, resulting in a rectangular shape for the probability distribution.

The given function f(x) = kx represents a probability density function (pdf) for a random variable x. In order for it to be a valid pdf, the integral of f(x) over its entire range must equal 1. Integrating f(x) = kx from 0 to 2, we get (k/2)x^2 evaluated from 0 to 2, which simplifies to k. Therefore, k must equal 1 for the function to be a valid pdf.

The given probabilities form a discrete probability distribution. To find p(x

The expected value, E(x), in a uniform distribution is equal to the average of the minimum and maximum values. In this case, since n = 19 is given as the value for the uniform distribution, the minimum and maximum values are both 19. Therefore, the expected value is (19 + 19) / 2 = 19. Since none of the answer choices match 19, the correct answer must be 10, which is not a valid explanation as it contradicts the correct answer.

A random variable is said to be continuous if it can take any value within a given interval. This means that there are no gaps or jumps between the possible values of the variable, and it can take on any real number within the specified range. In contrast, a discrete random variable can only take on specific, separate values, while a hybrid random variable is a combination of both continuous and discrete variables. Probability is a concept related to the likelihood of events occurring, but it is not directly related to whether a random variable is continuous or discrete.

The variance of a binomial distribution is given by the formula Var(x) = n * p * (1 - p). In this case, n = 20 and p = 1/10. Plugging in these values into the formula, we get Var(x) = 20 * (1/10) * (1 - 1/10) = 20 * (1/10) * (9/10) = 9/5.

A random variable which takes finite or countably infinite values is called a discrete random variable. This means that the possible outcomes of the variable can be listed and counted, such as the number of heads when flipping a coin or the number of cars passing by in a given hour. This is in contrast to a continuous random variable, which can take any value within a certain range, like the height of a person or the time it takes for a computer to process a task. A hybrid random variable does not exist, and "stop" is not a valid option for describing the type of random variable.

The c.d.f. (cumulative distribution function) represents the cumulative probability of a random variable taking on a value less than or equal to a given value. In this case, we are finding the c.d.f. of 3. Since the given probabilities are for specific values (p(1), p(3), p(5)), we need to find the cumulative probabilities up to 3. The probability of getting a value less than or equal to 3 is the sum of the probabilities of getting 1 and 3. Therefore, the c.d.f. of 3 is 3/5 + 3/10 = 9/10.

The variance of a random variable following an exponential distribution with parameter λ is equal to 1/λ^2. In this case, the parameter is 0.5, so the variance would be 1/(0.5)^2 = 4.