A simple random sample of size n = 81 is obtained from a population that is skewed right with µμ = 79 and o = 27. (a) Describe the sampling distribution of x. (b) What is P (x>83.05) ? (c) What is P (x≤73)? (d) What is P (75.25
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- The Central Limit Theorem says that the standard deviation of the sampling distribution of the sample means is (A exactly equal to the standard deviation. (B close to the population mean if the sample size is small close to the population standard deviation if the sample size is large. equal to the population standard deviation divided by the square root of the sample D size.Consider sample data with x = 8 and s = 4. (a) Compute the coefficient of variation. what percent? (b) Compute a 75% Chebyshev interval around the sample mean.Lower Limit =Upper Limit=Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 117 pounds and a standard deviation of 39.4 pounds. Random samples of size 15 are drawn from this population and the mean of each sample is determined. 4-0 0 = (Round to three decimal places as needed.) Sketch a graph of the sampling distribution. Choose the correct graph below. O A. A 96.7 117 137.3 Q O B. A 340.8 10.2 361.2 Q Q ww ^ A -106.8 10.2 127.2 86.5 117 147.5 O C. O D. Q
- What is the variance and standard deviation for the number of times the SAT is taken? x Frequency (f) fx f*x2= (f*x) x (x) (4) = (3)x(1) 5152727-(2547473)(2547473)/1518859=880035.3688/1518859=0.5794 (1) (2) (3)=(1)x(2) (4)=(3x1) Population Variance 1 721,769 721,769 721,769 2 601,325 1,202,650 2,405,300 3 166,736 500,208 1,500,624 4 22,299 89,196 356,784 5 6,730 33,650 168,250 Total 1,518,859 2,547,473 5,152,727 880035.3688 0.5794 n=1,518,859 ∑fx = 2,547,473 The population variance is 0.5794 The Standard Deviation? 0.76 Can you show me the steps to reach the standard deviation of .76 ?A normal distribution has μ = 30 and σ = 5. (a) Find the z score corresponding to x = 25. (Enter an exact number.) (b) Find the z score corresponding to x = 44. (Enter a number.) (c) Find the raw score corresponding to z = −3. (Enter an exact number.) (d) Find the raw score corresponding to z = 1.5. (Enter a number.)The thickness (in millimeters) of the coating applied to hard drives is one characteristic that determines the usefulness of the product. When no unusual circumstances are present, the thickness (x) has a normal distribution with a mean of 3 mm and a standard deviation of 0.05 mm. Suppose that the process will be monitored by selecting a random sample of 16 drives from each shift's production and determining x, the mean coating thickness for the sample. (a) Describe the sampling distribution of x for a random sample of size 16. The distribution of x is with mean mm and standard deviation 0.0125 mm. (b) When no unusual circumstances are present, we expect x to be within 2? x of 3 mm, the desired value. An x value farther from 3 mm than 2? x is interpreted as an indication of a problem that needs attention. Calculate 3 ± 2? x. 3 − 2? x = mm 3 + 2? x = mm (c) Referring to part (b), what is the probability that a sample mean will be outside 3 ± 2? x just by chance…
- Construct an interval estimate for the given parameter using the given sample statistic and margin of error.(a) For μ, using x̄=98 with margin of error 16.(b) For p1-p2, using p̂1-p̂2 = 0.165with margin of error 0.025Use the central limit theorem to find the mean and standard error of the mean of the indicated sampling distribution. Then sketch a graph of the sampling distribution. The per capita consumption of red meat by people in a country in a recent year was normally distributed, with a mean of 113 pounds and a standard deviation of 37.7 pounds. Random samples of size 20 are drawn from this population and the mean of each sample is determined.The heights of fully grown trees of a specific species are normally distributed, with a mean of 54.5 feet and a standard deviation of 5.75 feet. Random samples of size 19 are drawn from the population. Use the central limit theorem to find the mean and standard error of the sampling distribution. Then sketch a graph of the sampling distribution. The mean of the sampling distribution is u- %3D The standard error of the sampling distribution is o, = X. (Round to two decimal places as needed.) Choose the correct graph of the sampling distribution below. O A. В. OC. 2. 51.9 54.5 57.1 43.0 54.5 66.0
- The random variable x has a normal distribution with mean 50 and variance 9. Find the value of x, call it x0, such that: a) P(x ≤ xo) = 0.8413 b) P(x > xo) = 0.025 c) P(x > xo) = 0.95 d) P(41 ≤ x ≤ xo) = 0.8630Suppose that a random sample of size 64 is to be selected from a population with mean 50 and standard deviation 5. USE SALT (a) What are the mean and standard deviation of the sampling distribution of X? Mx μ- = 50 o = 0.625 X Describe the shape of the sampling distribution of x. The shape of the sampling distribution of x is approximately normal (b) What is the approximate probability that x will be within 0.5 of the population mean µ? (Round your answer to four decimal places.) (c) What is the approximate probability that x will differ from μ by more than 0.6? (Round your answer to four decimal places.)It is believed that the mean value of a variable measured for an entire population of individuals has value µ = 17 and that for the population the standard deviation is σ = 5.69. A sample of size n= 43 is to be taken from the population. According to the central limit theorem, the sampling distribution of the mean is Normal( 43 , 0.872 ) Normal( 17 , 5.692 ) Normal( 17 , 0.872 ) Normal( 43 , 5.692 )