Suppose a sample of 47 chocolate bars has a mean weight of 134.2 with standard deviation 7. Test the claim the mean weight is above 105.7 use a 5% level of significance
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A: given data normal distributionmean(μ) = 31.2standard deviation (σ) = 4.6for x = 39.5Z = ?
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A: X~N( μ , ?) μ=8.21 , ?=1.48 Z-score =( x - μ )/?
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A: Given: μ=7, σ=2, n=10 The corresponding z-value needed to be computed is,
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A: Given that. X~N( μ , ?^2 ) μ=210 , ?=16 Z-score =( x - μ )/?
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A: Given, chocolate chip cookies are normally distributed with mean μ = 1197 and σ = 142 We use the…
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A: percentage of values greater than 145 = 0.15% Explanation:Step 1: Define what the empirical rule…
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A: It is given that Mean, μ = 1281Standard deviation, σ = 118
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A: Given: Mean, μ = 8.33 Standard deviation, σ = 1.48 Concept: The empirical rule states that:…
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A: mu = 100 xbar = 102.5 sigma = 15 n = 144
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A: Answer: From the given data, mean(μ) = 160 standard deviation(σ) = 14
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A: Men mean =67.7 s = 3.5
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Suppose a sample of 47 chocolate bars has a
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- What is the forumula for standard deviationpoll used a sample of 100 randomly selected car owners. Within the sample, the mean time of ownership for a single car was 7.18 years with a standard deviation of 3.37 years. Test the claim by the owner of a large dealership that the mean time of ownership for all cars is less than 7.5 years. Use a 0.05 significance level. Find the Z score and P value and if left-tailed or right-tailed. Also to reject or not too.Total blood volume (in ml) per body weight is important in medical research. For healthy adults, the red blood cell volume mean is about 34 ml/kg. Red blood cell that is too low or too high can indicate a medical problem. Suppose that Roger has 20 blood tests and the mean is 35.4 ml/kg. Suppose that red blood cell volume follows normal distribution with standard deviation 2.34 ml/kg. With 5% significance level, test that the mean blood cell volume is different from 28ml/kg. Which test to use? a.) None of the answer choices b.) T-test for the Mean c.) T-test for Proportion d.) Z-test for the Mean e.) Z-test for Proportion
- Experimental components for use in aircraft engines were tested to destruction under extreme conditions. The survival times, in days, of ten components were tested. The mean survival time was determined to be 445.5 days with a standard deviation of 120 days. The experimental components manufactured by the company report that the mean survival time is 430 days. Is there a significant difference between the sample and the claim? Evaluated on a TI 83 calculatorfind and interpret the z-score for the darta value given/ the value 250 in a dataset with mean 208 amd standard deviation 21.Suppose that IQ scores have a bell-shaped distribution with a mean of 99 and a standard deviation of 17. Using the empirical rule, what percentage of IQ scores are less than 133? Please do not round your answer.
- what is the minimum pregnancy length that can be in the top 8% of pregnancy given a mean of 265 and a standard deviation of 12The number of orders received daily by an online vendor of used CDs is normally distributed with mean 260 and standard deviation 20. The company has to hire extra help or pay overtime on those days when the number of orders received is 310 or higher. What percentage of days will the company have to hire extra help or pay overtime? Click the icon to view the standard normal distribution table. The company will have to hire extra help or pay overtime on % of days.The number of orders received daily by an online vendor of used CDs is normally distributed with mean 240 and standard deviation 16. The company has to hire extra help or pay overtime on those days when the number of orders received is 264 or higher. What percentage of days will the company have to hire extra help or pay overtime? Click the icon to view the standard normal distribution table. The company will have to hire extra help or pay overtime on % of days.
