Given the probability distribution of the random variable X below, compute for the variance and standard deviation.
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A: It is an important part of statistics. It is widely used.
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A: Answer: From the given data, p = 0.37 and q = 1 - p = 1 - 0.37 = 0.63 sample size ( n )= 22
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A: Answer: From the given data, The parameters of a binomial distribution are given as, n = 32 p = 0.26
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A:
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A: For aa standard normal variable, mean is 0 and standard deviation is 1.
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A: Answer: From the given data, P = 0.72 Sample size(n) = 26
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A:
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Q: Find the mean and variance for the binomial distribution n= 50 and p= 0.25
A: Answer: From the given data, Using binomial distribution with parameters, n = 50 p = 0.25
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A: The question is about discrete prob. dist. Given :
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A: we hvae to find the probability.:_ The probability that Z is less than −0.25.
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A: The mean is calculated as: μ=∑x·p(x)=2×0.2+6×0.5+8×0.3=5.8
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A: Answer: From the given data, For the binomial distribution, P = 0.29 n = 13
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A: Answer: From the given data, Sample size(n) = 12 p = 0.65 X follows binomial distribution (n,p)
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Given the probability distribution of the random variable X below, compute for the variance and standard deviation.
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- Assume x is a normally distributed random variable with mean = 50 and variance = 3. Find the value of the random variable x0 when P(x ≤ x0) = .8413 A. 33 B. 49 C. 53 D. 16Find the mean and variance for the binomial distribution P=0.33 and n=26Write the probability density function of a normal random variable and a standard normal random variable. If the annual rainfall in Cape Town is normally distributed with mean 20.2 inches and standard deviation 3.6 inches. Find the probability that the sum of the next five years’ rainfall exceeds 110 inches.
- The random variable X, representing the number of errors per 100 lines of software code, has the following probability distribution: x 2 3 4 5 6 f(x) 0.01 0.25 0.4 0.3 Find the standard deviation of X.Fill in the blanks: Let X be the random variable with the following distribution: х: -2 3 5 P (X = x) : 0.5 0.3 0.2 Determine the variance of X.17 percent of U.S. employees who are late for work blame oversleeping. You randomly select four U.S. employees who are late for work and ask them whether they blame oversleeping. The random variable represents the number of U.S. employees who are late for work and blame oversleeping. Find the mean of the binomial distribution. μ=
- We are examining farmer production in marginal land. A sample of 100 farmers produce this crop with mean of 100 bushels per acre. Assume a normal distribution. Q1: What is the probability an individual farmer in this population produces 103 or less bushels of corn?Q2: What is the probability the mean bushels of corn produced is 103 or less bushels of corn?If z is a standard normal random variable, find the probability.The probability that z lies above 1.12Find the probability for a normal distribution p(-0.80
- Find the mean and variance of the binomial Distribution P=0.38, n=5A snack company makes packages of grapes and honeydew slices. The weight of an individual package of grapes, G, is approximately Normally distributed with a mean of 3.25 ounces and a standard deviation of 0.91 ounces. The weight of an individual package of honeydew slices, H, is approximately Normally distributed with a mean of 4.51 ounces and a standard deviation of 2.02 ounces. Assume G and H are independent random variables. Let D = G – H. What is the probability that a randomly selected package of grapes weighs more than a randomly selected package of honeydew slices? 0.127 0.230 0.284 0.334