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- 5. The graph of the chi-square distribution with 5 degrees of freedom is shown in Fig. 1. Find the values xỉ and x for which (a) the shaded area on the right = 0.05 (b) the total shaded area = 0.05 (c) the shaded area on the left = 0.10 (d) the shaded area on the right = 0.01 Figure 1Q. 2 Describe the method of Moving averages for estimating the trend in a time seriesA population is normally distributed with mean 18.6 and standard deviation 1.4. (a) Find the intervals representing one, two, and three standard deviations of the mean. one standard deviation (smaller value) (larger value) two standard deviations (smaller value) (larger value) three standard deviations (smaller value) (larger value) (b) What percent of the data lies in each of the intervals in part (a)? (Round your answers to two decimal places.) one standard deviation % two standard deviations % three standard deviations %
- A population is normally distributed with mean 19 and standard deviation 1.1. (a) Find the intervals representing one, two, and three standard deviations of the mean. one standard deviation (smaller value) (larger value) two standard deviations (smaller value) (larger value) three standard deviations (smaller value) (larger value) (b) What percent of the data lies in each of the intervals in part (a)? (Round your answers to two decimal places.) one standard deviation % two standard deviations % three standard deviations % (c) Draw a sketch of the bell curve.ats 2. In the mid-19th century, Sir Francis Galton demonstrated, through extensive data collection, that many human characteristics such as height and IQ are distributed throughout the entire population in accordance with the normal distribution. Galton recorded the heights of 8585 men in Great Britain. This forms a normal distribution with a mean of 67 inches and standard deviation of 2.5 inches. (a) Graph this normal distribution and label the x-axis. (b) What proportion of these men were between 66 and 68 inches tall? belool vin (c) What fraction of the adult male population in Great Britain was within one standard deviation of the mean height? [Use the normal distribution function in Excel, not the Empirical Rule] (d) What height corresponds to the 95th percentile in the Galton study?A vertical line is drawn through a normal distribution so that 47.5% of the distribution is located between the line and the mean. The line is drawn at z = 1,96 or at z = -1.96. Question 13 options: True or False
- Withdrawal symptoms may occur when a person using a painkiller suddenly stops using it. For a special type of painkiller, withdrawal symptoms occur in 4% of the cases. Consider a random sample of 2400 people who have stopped using the painkiller. Answer the following. (If necessary, consult a list of formulas.) (a) Find the mean of p, where p is the proportion of people in the sample who experience withdrawal symptoms. (b) Find the standard deviation of (c) Compute an approximation for P(p <0.05), which is the probability that fewer than 5% of those sampled experience withdrawal symptoms. Round your answer to four decimal places.suppose the mean is 8 and 1/3 and the population variance is 6 and 8/9 and a>9 x 5 9 a frequency 3 f 2 a=? f=?A population is normally distributed with mean 16.3 and standard deviation 1.4. (a) Find the intervals representing one, two, and three standard deviations of the mean. one standard deviation (smaller value) (larger value) two standard deviations (smaller value) (larger value) three standard deviations (smaller value) (larger value) (b) What percent of the data lies in each of the intervals in part (a)? (Round your answers to two decimal places.) one standard deviation % two standard deviations % three standard deviations %
- QUESTION 14 2. An equity analyst wants to verify whether there is a variation in the mean rate of return for 7 types of equity. The analyst selected a sample of 135 and conducted an ANOVA test that is shown as follows. Variation SS df MS F Between 218 (Treatment) Within (Error) ? Total 717 The Within (Error) degree of freedom is equal to?SITUATION 6 (a) The elongation of a steel bar under a particular load has been established to be normally distributed with mean of 0.05 inch and standard deviation of 0.01 inch. (a.1) Find the probability that the elongation of a steel bar is below 0.15 inch. (a.2) Find the probability that the elongation of a steel bar is between 0.025 and 0.050 inch.