Binomial distribution calculati If n=5 and p=0.5, find P(x=1) Give at least 4 decimal places
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A: As per guideline expert have to answer first three subparts only.
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A: Given information rate=1/12
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A: We have given that, (a) n = 30, π = 0.50 (b) n = 50, π = 0.20 (c) n = 100, π = 0.10…
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A: Suppose is the population standard deviation of the ages of seniors at the college.
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A: Reviewing information, Cdf F~exp(λ=1)Cdf G~exp(λ=2)
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A: X(net gain) Probabiliy-8 .0983692…
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A: The obtained multiple regression equation is Weight=1.0+85.3*Sex+2.9*Age.
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A: Given, Sample size, n = 38Sample mean, = 215Population standard deviation, σ = 24Confidence level,…
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A: Given: Population mean μ=0 Population standard deviation σ=2 X ~ ln N(0, 22) lnX ~ N (0, 22)
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A: n=90p=0.73
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A: Width = 10 Bottom value = 15
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A: We have given that Sample size (n) = 29Sample standard deviation(s) = 2.2
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A: 1) In the given situation, X follows Binomial distribution with n=5 and p=θ. PX=x|θ=5xθx1-θ5-x,…
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A: Hi! Thank you for the question, as per the honor code, we are allowed to answer three sub-parts at a…
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Q: None
A: To find the square root of the standard deviation, for a binomial distribution, we can use the…
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A: Given X~exp(β=1)
Q: Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with…
A: Given : Population mean , μ = 29.7 kg Population standard deviation , σ = 4.4 kg Let x be the weight…
Q: probability to be no greater than 0.20 when μ sume that σ = 4. = 12? As-
A: Step 1: To understand test outputs, we need to understand Z-values in the context of hypothesis…
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A: Given : Mean =29 Standard Deviation=5
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A: From the given informations: fx=120, 25<x<45
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- NT #4 An unknown distribution has a mean of 40 and a standard deviation of 8. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is greater than 2,400. NT #5Normal distribution mean=0 standard deviation =1 p(z<-0.21)mean - 490 standard deviation 40 nt find the chance of selecting ti 450 or less Determine the minimum number that should be sampled such that the probability associated with selecting a sample with a mean greater than 510 is 0.025. 4. ele
- http Suppose the stock price is 40 and we need to price a call option with a strike of 45 maturing in 4 months. The stock is not expected to pay dividends. The continuously-compounded riskfree rate is 3%/year, the mean return on the stock is 7%/year, and the standard deviation of the stock return is 40%/year.ages of a variation than ages of females in the general population, so let o = 18.4 years for the sample size calculation. How many female statistics student ages must be obtained in order to estimate the mean age of all female statistics students? Assume that we want 98% confidence that the sample mean is within one-half year of the population mean. Does it seem reasonable to assume that the ages of female statistics students have less variation than ages of females in the general population? The group of 137 randomly selected adult females have a standard deviation of 18.4 years. Assume that the ages of female statistics students have less The required sample size is. (Round up to the nearest whole number as needed.)Finding Probabilities in Normal Distributions Weekly salaries of college students in a large city are normally distributed with a mean of 540 dollars and a standard deviation of 45 dollars. Find the probability that a randomly selected weekly salary in this distribution is less than 441 dollars. P(X < 441) = Enter your answer as a number accurate to 3 decimal places. Answers obtained using exact z-scores or z- scores rounded to 3 decimal places are accepted.
- Group I Given N1 = 20 Sigma(X – 8) = 10 Sigma (X – 8) ^ 2 = 152 Group II N2=30 Sigma (Y-10)=-15 Sigma(Y-10)^2=300 Find the Arithmetic mean and standard deviation of each group and the combined Arithmetic mean, standard deviation and the overall coefficient of variation.Cumulative probabilitySolve it quickly please
- Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean ? = 26.6 kilograms and standard deviation ? = 3.4 kilograms. Let x be the weight of a fawn in kilograms. Convert the following x intervals to z intervals. (Round your answers to two decimal places.) (a) x < 30 z < _______ (b) 19 < x _______ < z (c) 32 < x < 35 ________ < z <_________Q life time of led follow exponential distribution with mean life of 5000 burning hours. if one such LED is selected at random from a big lot, what is the probability that it will last for 1. at least 4000 hours 2. almost 10000 hours Also state the standard deviation of the life time distribution. Given e raised to -0.8 is 0.4493 and e raised to -2 is 0.13533 and 4 please
