nLet x be a continuous random variable that has a normal distribution with u = 48 and o = 8. Assuming< 0.05, find theprobability that the sample mean, x, for a random sample of 16 taken from this population will be between 49.62 and 53.00.Round your answer to four decimal places.P(49.62 < I < 53.00) =i

# Given: random variable x follow normal distribution with mean=48 and standard deviations =8…

Answered: Let x be a continuous random variable…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The random variable X has a uniform distribution with values between 10 and 18. What is the mean and standard deviation of X? (Round your answer to three decimal places.) Select the correct answer below: mean is 14; standard deviation is 43‾√3≈2.309 mean is 10; standard deviation is 43‾√3≈2.309 mean is 14; standard deviation is 83≈2.667 mean is 10; standard deviation is 46‾√3≈3.266 mean is 10; standard deviation is 83≈2.667 mean is 14; standard deviation is 46‾√3≈3.266
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T3.6
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Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in mg/100 ml). 95 86 82 105 97 112 86 89 The sample mean is x ≈ 94.0. Let x be a random variable representing glucose readings taken from Gentle Ben. We may assume that x has a normal distribution, and we know from past experience that σ= 12.5. The mean glucose level for horses should be μ= 85 mg/100 ml.† Do these data indicate that Gentle Ben has an overall average glucose level higher than 85? Use ? = 0.05. (a) Compute the z value of the sample statistic. (Round your answer to two decimal places.) (b) Find (or estimate) the P-value. (Round your answer to four decimal places.)
The random variable X has a uniform distribution with values between 14 and 22. What is the mean and standard deviation of X? (Round your answer to three decimal places.) Select the correct answer below: mean is 18; standard deviation is 83≈2.667 mean is 14; standard deviation is 46–√3≈3.266 mean is 18; standard deviation is 43–√3≈2.309 mean is 14; standard deviation is 83≈2.667 mean is 14; standard deviation is 43–√3≈2.309 mean is 18; standard deviation is 46–√3≈3.266
Gentle Ben is a Morgan horse at a Colorado dude ranch. Over the past 8 weeks, a veterinarian took the following glucose readings from this horse (in mg/100 ml). 95 86 82 The sample mean is x ≈ 93.8. Let x be a random variable representing glucose readings taken from Gentle Ben. We may assume that x has a normal distribution, and we know from past experience that o = 12.5. The mean glucose level for horses should be μ = 85 mg/100 ml. + Do these data indicate that Gentle Ben has an overall average glucose level higher than 85? Use a = 0.05. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? H₁:μ >85; H₁: μ = 85; right-tailed 107 99 108 82 91 ⒸH₁: μ = 85; H₁: μ 85; two-tailed H₁: μ = 85; H₁: >85; right-tailed H₁: μ = 85; H₁₂ : μ< 85; left-tailed (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The Student's t, since n is large with unknown…
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A normal distribution has a mean of μ = 36 with σ = 4. What is the probability of randomly getting a score between 30 and 38? Group of answer choices 0.2583 0.3753 0.7417 0.6247

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