Let z denote a variable that has a standard normal distribution. Determine the value z* to satisfy the following conditions. (Round all answers to two decimal places.) (a) P(z < z*) = 0.0256 z* =
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Let z denote a variable that has a standard
z* =
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- ANSWER COMPLETELY FOR UPVOTEA. P(t10>1.75), where t10 has a t distribution with 10 degrees of freedom(a) Let Z follow a standard normal distribution. i. Find the equi-tailed 95% probability interval, i.e. find a > 0 such that P(Z € (-a, a)) = 0.95 and express your result in terms of the inverse function -1 of the standard normal cdf (note that the inverse function satisfies -((x)) = x for all r E R). Finally write down the value of a to three significant digits. ii. Instead of an equi-tailed interval, consider the interval (-a, b) where a, b >0 are such that P(ZE (-a, b)) = 0.95. Express b as a function b(a) of a. It may help to express your results in terms of -1 and . Show that the derivative of the length I of the interval, I = b(a) + a, is given by dl = 1 da (a) (-(0.95+ (-a))' where o denotes the standard normal pdf. Hence obtain a candidate value of a for which the length of the interval is minimal by equating this derivative to zero. You do not need to show that this candidate value is an actual minimizer.
- Central High School believes their students have unusually high SAT scores on average. The school has 165 students.Based on national data, the average SAT score is 1067 with a population standard deviation of 207. Assume SAT scores are normally distributed. Let X be the random variable representing the mean SAT scores for groups of 165 randomly selected students.a. Fill in the blank, rounding your answers to 2 decimal places if needed. According to the Central Limit Theorem, X is approximately normal with a mean of and a standard error of the mean .b. Find the z-score associated to a sample with a mean of 1109, using the sampling distribution. Round your answer to two decimal places. c. Find the probability that a randomly selected sample of 165 students has a mean SAT score higher than 1109. Round your answer to 4 decimal places. d. Central High School finds that for their students, the average SAT score is 1109. Are they justified in saying their students perform unusually well on…Let T represent the lifetime in years of a part which follows a Weibull distribution with shape 2 and scale 5. (a) What is E(T )? Make sure to simplify the gamma function in terms of pi.(b) What is V(T )? Make sure to simplify the gamma function in terms of pi.STAT 202: Individual Assignment x>0 1. Given the Gamma distribution as f(x) =- elsewhere Find the mean and variance of the distribution. 2a. Suppose that a large conference room for a certain hotel can be reserved for no more than 4 hours. However, the use of the conference room is such that both long and short conferences occur quite often. In fact, it can be assumed that length X of a conference has a uniform distribution on the interval [0, 4]. i. Determine the probability density function ii. What is the probability that any given conference last at least 3 hours? b. A cement wholesale distributor has a large store that hold fixed supplies and are filled every Sunday. Of interest to the wholesaler is the proportion of this supply that is sold during the week. Over many weeks of observation, the distributor found out that this proportion could be modeled by a beta distribution with a = 4 and ß = 2 . Find the probability that the wholesaler will sell at least 90% of his stock in a…
- 6.9 To enter the cumulative normal distribution table Þ(Z), calculate Z and Z at the spec limits. upper lower Þ(2) = 0.9772 and Þ(-2) = 0.0228, the area to the right of the USL plus to the left of the LSL is 0.0456.Let the test statistic T have a t distribution when Ho is true. Give the P-value for each of the following situations. (Round your answers to three decimal places.) USE SALT (a) Ha: μ> Mo, df = 15, t = 2.864 P-value = 0.983 X (b) Ha: μ< μo, n = 24, t = 2.234 P-value = (c) H₂:μμ₁, n = 31, t = -1.691 or t = 1.691 P-value =Consider the normal distribution represented by the normal curve in the figure to the right. (Assume the two labeled values are equidistant from the mean.) Complete parts (a) and (b) below. 95% of the data 73.2 cm 96.8 cm (a) Find the mean μ and standard deviation σ of the distribution. με cm (Simplify your answer.) σ = cm (Simplify your answer.) (b) Find the first quartile Q1 and the third quartile Q3 of the distribution. ༠ ༤ cm (Round to the nearest tenth as needed.) Q3≈ cm (Round to the nearest tenth as needed.)
- Let x = red blood cell (RBC) count in millions per cubic millimeter of whole blood. For healthy females, x has an approximately normal distribution with mean ? = 5.6 and standard deviation ? = 0.4. (a) Convert the x interval, 4.5 < x, to a z interval. (Round your answer to two decimal places.) (b) Convert the x interval, x < 4.2, to a z interval. (Round your answer to two decimal places.) (c) Convert the x interval, 4.0 < x < 5.5, to a z interval. (Round your answers to two decimal places.)Conclusion: Fail to reject Null hypothesis or reject Null hypothesisDetermine to 3 decimal places the value of x² for 25 degrees of freedom and an area of 0.990 in the left tail of the chi-square distribution curve. x² = i
