Suppose a random sample size of 60 is taken from a population with p= 0.51. The standard error of the sample proportion is %3D 0.0645 0.0495 0.0891 0.0100
Q: in a random sample of 92 trout, the average weight is 8 pounds, with standard Heviation(s) = 0.4,…
A: Answer: From the given data, Sample size(n) = 92 mean(x¯) = 8 Standard deviation(S) = 0.4
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A: Answer: From the given data, Sample size(n) = 100 mean (x¯) = 10 standard deviation(s) = 0.6
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A: Standard normal distributionZ~N(0,1)mean = 0, standard deviation = 1Proportion of area between mean…
Q: In a random sample of 80 trout, the average weight is 8 pounds, with standard deviation (s) = 0.9,…
A: Answer: From the given data, Sample size(n) = 80 mean = 8 Standard deviation(s) = 0.9
Q: A group of biologists investigated a species of worms whose length can be characterised by a normal…
A: The objective of this question is to find the length of the worms such that only 1% of the specimens…
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A: Solution : Given : Confidence level = 86.12% and population variance is known. When population…
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A: Given that, The z-score value is -1.2.
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A: From the provided information,Population standard deviation = 1.3Sample size (n) = 12
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A: In statistical inference of two samples ,the margin of error is obtained by multiplying the standard…
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A: List of formulae : Mean of sampling distribution of sample proportion ( p^ ) : μp^ = p Standard…
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A: given the random variable X follows normal distribution mean(μ) =0variance =800we know the formula…
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A: Solution: From the given information, X follows normal distribution with mean µ=59 and a standard…
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A: For aa standard normal variable, mean is 0 and standard deviation is 1.
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A: Given sample of size n=81 mean=88 standard deviation =27.
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A: Given data, p=0.53 n=105
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A: The variance of the sample can be calculated as:
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A: given data normal distribution μ = 24σ = 3 to be in top 14% minimum required score = ?
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Q: Samples of size n = 9 are selected from a normal population with µ = 80 with σ = 18. What is the…
A: Given information n = 9 µ = 80 σ = 18
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A: n1=282, n2=287
Q: For the case where the population ratio to population is p = 0.08, what is the probability of…
A: It is an important part of statistics. It is widely used.
Q: What are the mean and the standard deviation for the sampling distribution of the sample proportion?
A: Given that, Sample size (n) = 100 Population proportion (p) = 0.60 q = 1-p = 1-0.60 = 0.40 To find…
Q: A population of N = 6 scores has ΣX = 42. What is the population mean?
A: Given ,Population of N=6 scoresand Σx =42
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Q: for a normal distribution with a population mean of 74 and a standard deviation of 8, what is the…
A: given data normal distributionμ = 74σ = 8to saperate top 60% x = ?
Q: For a sample of n = 11 scores with SS = 440. What is the sample variance and the estimated standard…
A: Sample Variance s2 = SS/n-1 Estimated standard error sM = s/sqrt(n).
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- A group of biologists investigated a species of worms whose length can be characterised by a normal random variable with mean 3.10cm and standard deviation 0.06cm. What is the length in cm such that 1% of the specimens in that species exceed that length? (2 d.p.)?A toy manufacturer wants to know how many new toys children buy each year. Assume a previous study found the variance to be 1.44. She thinks the mean is 6.7 toys per year. What is the minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 85% level of confidence? Round your answer up to the next integer.In a random sample of 87 trout, the average weight is 6 pounds, with standard deviation(s) = 0.8, Find the standard error of the sample mean.
- Determine the sample mean, sample variance, and sample standard deviationFind the value of variance in the Binomial distribution where there are 5 samples and p=0.3.The amount of trash generated by US households (in lbs per day) is normally distributed. You take a sample of 10 people. The sample average is xbar=10.91 lbs, and the sample standard deviation is s=4.736. What is the 99% Conf. Int. for the mean? Group of answer choices a. 6.04 to 15.78 b. 8.33 to 13.49 c. 8.09 to 13.73 d. 7.74 to 14.08
- In a simple random sample size of 89, there were 40 individuals in the category of interest. It is desired to test —- H0:P=0.46 VS H1:PGiven a Normal Distribution with standard error of 11 from a population with a mean of 135 and a standard deviation of 66, find the sample size (i.e. how many people or items were sampled from the population?). n =A CNC (computer numerical control) machine produces iron automobile crankshafts. Samples are measured and the inner diameter are found to be 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. For all computations, assume an approximately normal distribution. The sample mean and standard deviation for the given data are = 1.0056, and s =0.0246. Find a 99% confidence interval on the mean of diameter. Compute a 99% prediction interval on a measured diameter of a single crankshaft piece taken from the machine. Find the 99% tolerance limits that will contain most of the metal pieces produced by the CNC machine. O a. CI on the mean: 0.9781 <µ <1.0331. Prediction Interval: 0.9186 s Xp+151.0926. Tolerance Interval (0.8937 and 1.1175) O b. Insufficient data to compute. Missing the sample size n Oc CIon the mean: 0.9781 <µ<1.0331. Prediction Interval: 0.9186 s Xp+151.0926. Tolerance Interval (0.8937 and 1.1175). Insufficient data to compute for the Tolerance Interval Od. CI on…
- A machine produces glass with a nominal thickness of 4mm. In fact the thickness is a Normal random variable with mean 4.168mm and standard deviation 0.12mm. The thickness in mm which 5% of panes exceed is (2 d.p.)?answer in 30minsUsing an evenly distributed data population of 8,000 samples, how many Would fall between a Z-score of -0.5 and +0.5? (blank)