Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n 125, p 0,36 The mean, u., is (Round to the nearest tenth as needed.)
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- Internet Service Providers The amount that households pay service providers for access to the Internet varies quite a bit, but the mean monthly fee is $50 and the standard deviation is $20. The distribution is not Normal: many households pay a low rate as part of a bundle with phone or television service, but some pay much more for Internet only or for faster connections. A sample survey asks an SRS of 50 households with Internet access how much they pay. Let x be the mean amount paid. 1. Explain why you can't determine the probability that the amount a randomly selected household pays for access to the Internet exceeds $55. 2. What are the mean and standard deviation of the sampling distribution of x ? 3. What is the shape of the sampling distribution of x ? Justify your answer. 4. Find the probability that the average fee paid by the sample of households exceeds $55.Find the range, the standard deviation, and the variance for the given samples. Round non-integer results to the nearest tenth. 6, 12, 13, 14, 15, 18, 19 range standard deviation varianceWhich of the following statements are true? It is okay to use the median to estimate the mean since both are a measure of the center of a distribution. It is okay to use the standard deviation to estimate the IQR since both are measures of variability. Point estimates based on a sample are sometimes far from a parameter's value. i only ii only iii only i and ii i and iii i, ii, and iii
- Find the range, the standard deviation, and the variance for the given samples. Round non-integer results to the nearest tenth. 3, 5, 9, 15, 18, 21, 25 range standard deviation varianceFind the sample variance and the standard deviation for the following sample. Round the answers to at least two decimal places as needed. 16 30 22 18 11 10 Send data to Excel The sample variance is The sample standard deviation isQ13. Derive mean of gamma distribution.
- The purpose of this assignment is to check that you are able to properly use the statistical keys on your calculator to find the mean and standard deviation. The data below represents the number of siblings that each student has in a random sample of 10. Enter the data into your calculator, then find the mean and standard deviation. Express each to the tenths place. (to one decimal place) 2 6 3 1 1 6 2 1 3 4 A. sample mean x̄ ____________ B. sample standard deviation s ____________You may change one number in the table if needed to make it work x -2 2 4 1 P(x=x) 0.21 -0.03 0.32 0.18 Give a 2-decimal answer for each: Mean Variance Standard DeviationFind the variance of the binomial distribution for which n=800 and p=0.93. Round the answer to the nearest hundredth. OA. 52.08 O B. 32.54 OC. 7.22 D. 744.00
- Find the margin of error for the given values of c, s, and n. c=0.80, s = 6, n = 10 Click the icon to view the t-distribution table. The margin of error is (Round to three decimal places as needed.)t-Distribution Table d.f. 123456789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Level of confidence, c 0.80 0.90 0.95 0.98 0.10 0.05 0.025 0.01 0.10 0.05 0.02 0.20 3.078 6.314 12.706 31.821 1.886 2.920 4.303 6.965 1.638 2.353 3.182 4.541 1.533 2.132 2.776 3.747 1.476 2.015 2.571 3.365 1.440 1.943 2.447 3.143 1.415 1.895 2.365 2.998 1.397 1.860 2.306 2.896 1.383 1.833 2.262 2.821 1.372 1.812 2.228 2.764 2.201 2.718 2.179 2.681 3.055 2.160 2.650 3.012 2.624 2.977 2.131 2.602 2.947 2.583 2.921 2.898 2.567 2.552 2.878 One tail, a Two tails, a 0.99 0.005 0.01 d.f. 63.657 1 9.925 2 5.841 3 4.604 4 4.032 5 3.707 6 3.499 7 3.355 8 3.250 9 3.169 10 3.106 11 12 13 14 15 16 17 18 2.539 2.861 19 1.363 1.796 1.356 1.782 1.350 1.771 1.345 1.761 2.145 1.341 1.753 1.337 1.746 2.120 1.333 1.740 2.110 1.330 1.734 2.101 1.328 1.729 2.093 2.086 2.528 1.325 1.725 2.845 20 1.323 1.721 2.080 2.518 2.831 21 1.321 1.717 2.074 2.508 2.819 22 1.319 1.714 2.069 2.500…Find the standard deviation for the group of data items. 5, 6, 7, 7, 8, 9 The standard deviation is (Simplify your answer. Round to two decimal places as needed.)
