A lot of 75 washers contains 5 defectives whose variability in thickness is unacceptably large. A sample of 10 washers is selected at random without replacement. Calculate the standard deviation of (x).
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A lot of 75 washers contains 5 defectives whose variability in thickness is unacceptably large. A sample of 10 washers is selected at random without replacement. Calculate the standard deviation of (x).
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- Suppose that a researcher is interested in estimating the mean systolic blood pressure, u, of executives of major corporations. He plans to use the blood pressures of a random sample of executives of major corporations to estimate u. Assuming that the standard deviation of the population of systolic blood pressures of executives of major corporations is 26 mm Hg, what is the minimum sample size needed for the researcher to be 99% confident that his estimate is within 4 mm Hg of u? Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements). (If necessary, consult a list of formulas.)The head of the quality department of a battery factory needs to estimate the average lifetime of a large delivery of “double A” batteries made at the factory. The standard deviation of lifetimes is known to be 100 hours. A random sample of 64 “double A” batteries results in a sample mean lifetime of 350 hours. How large a sample size is needed if we wish to be 99% confident that the mean lifetime for the entire batteries will be within +350 minutes?CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 271 dollars. You take a simple random sample of 96 auto insurance policies.Find the probability that a single randomly selected value is less than 969 dollars.P(x<969)=P(x<969)= Find the probability that a sample of size n=96n=96 is randomly selected with a mean less than 969 dollars.P(¯x<969)=P(x¯<969)= Enter your answers as numbers accurate to 4 decimal places.
- CNNBC recently reported that the mean annual cost of auto insurance is 1020 dollars. Assume the standard deviation is 201 dollars. You take a simple random sample of 84 auto insurance policies.Find the probability that a single randomly selected value is at least 981 dollars.P(X > 981) = Find the probability that a sample of size n=84n=84 is randomly selected with a mean that is at least 981 dollars.P(M > 981) = Enter your answers as numbers accurate to 4 decimal places.Suppose that a researcher is interested in estimating the mean systolic blood pressure, μ, of executives of major corporations. He plans to use the blood pressures of a random sample of executives of major corporations to estimate μ. Assuming that the standard deviation of the population of systolic blood pressures of executives of major corporations is 26mm Hg, what is the minimum sample size needed for the researcher to be 95% confident that his estimate is within 5mm Hg of μ? Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements).A certain type of thread is manufactured wen a mean tensile strength of 75.3 klograms and a standard deviation of 4.8 kilograms. How is the variance of the sample mean changed when the sample size (a)increased from 64 to 900 ? (b) decreased from 256 to 36 ?
- A random sample of 51 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.01 years, with sample standard deviation s = 0.76 years. However, it is thought that the overall population mean age of coyotes is u = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H 1.75 yr; Hi: и 1.75 yr; Hi: и %3D 1.75 yr II Но: и 3 1.75 уr; Hi: и > 1.75 yr Ho: H = 1.75 yr; H1: µ # 1.75 yr (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The standard normal, since the sample size is large and o is known. The Student's t, since the sample size is large and o is known. The standard normal, since the sample size is large and o is unknown. O The Student's t, since the sample size is large and o is unknown. What is the value of the…Have American males (AMs) gotten heavier over the last 20 years? A random sample 77 AMs in 2019 had a mean weight of x = 189.030 pounds. A random sample 93 AMs in 1999 had a mean weight of y = 184.795 pounds. It is recognized that the true standard deviation of 2019 AMs weights is σx = 14.04 pounds while it is recognized that the true standard deviation of 1999 AMs weights is σy = 10.03 pounds. The true (unknown) mean of 2019 AMs weights is μx pounds, while the true (unknown) mean of 1999 AMs weights is μy pounds. Weights are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation 2019 Data (X) 77 189.030 14.04 1999 Data (Y) 93 184.795 10.03 d) Calculate the standard deviation of X - Y? . e) If we wish to create an 98% confidence interval for μx-μy then what is the z critical value used? f)Create an 98% confidence interval for μx-μy .CNNBC recently reported that the mean annual cost of auto insurance is 973 dollars. Assume the standard deviation is 234 dollars. You take a simple random sample of 61 auto insurance policies. Find the probability that a single randomly selected value is less than 991 dollars.P(X < 991) = Find the probability that a sample of size n=61n=61 is randomly selected with a mean less than 991 dollars.P(M < 991) =
