solve these problems

I have three errands to take care of in the Administration Building. Let X; = the time that it takes for the ith errand (i = 1, 2, 3), and let X, = the total time in minutes that I spend walking to and from the building and between each errand. Suppose the X/'s are independent, and normally distributed, with the following means and standard deviations: X; = 0, =3 | plan to leave my office at precisely 10:00 A.M. and wish to post a note on my door that reads, "I will return by t A.M." What time t should I write down if I want the probability of my arriving after f to be .01?In Chapter 3, we defined a negative binomial rv as the number of failures that occur before the th success in a sequence of independent and identical success/failure trials. The probability mass function (pmf) of X is nb(x; r, p) = xtrol p( - py x = 0, 1, 2, . .. X a. Suppose that / 2 2. Show that is an unbiased estimator for p. [Hint: Write out E( ) and cancel x + r - 1 inside the sum.] b. A reporter wishing to interview five individuals who support a certain candidate begins asking people whether (S) or not (F) they support the candidate. If the sequence of responses is SFFSFFFSSS, estimate p = the true proportion who support the candidate.Reconsider the minicomputer component lifetimes X and Y as described in Exercise 12. Determine E(XY). What can be said about Cov(X, Y) and p? Reference exercise 12 Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) xe *(1+>) x 2 0 andy 2 0 0 otherwise a. What is the probability that the lifetime X of the first component exceeds 3? b. What are the marginal pdf's of X and Y? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds 3?The following observations are lifetimes (days) subsequent to diagnosis for individuals suffering from blood cancer ("A Goodness of Fit Approach to the Class of Life Distributions with Unknown Age," Quality and Reliability Engr. Intl., 2012: 761-766): 115 181 255 418 441 461 516 739 743 789 807 865 924 983 1025 1062 1063 1165 1191 1222 1222 1251 1277 1290 1357 1369 1408 1455 1478 1519 1578 1579 1599 1603 1605 1696 1735 1799 1815 1852 1899 1925 1965 a. Can a confidanas Inthe. I a. Can a confidence interval for true average calculated without assuming anything lifetime be about the nature of the lifetime distribution? Explain your reasoning. [Note: A normal probability plot of the data exhibits a reasonably linear pattern.] b. Calculate and interpret a confidence interval with a 99% confidence level for true average lifetime. [Hint: : = 1191.6 and s = 506.64According to the article "Fatigue Testing of Condoms" (Polymer Testing, 2009: 567-571), "tests currently used for condoms are surrogates for the challenges they face in use," including a test for holes, an inflation test, a package seal test, and tests of dimensions and lubricant quality (all fertile territory for the use of statistical methodology!). The investigators developed a new test that adds cyclic strain to a level well below breakage and determines the number of cycles to break. A sample of 20 condoms of one particular type resulted in a sample mean number of 1584 and a sample standard deviation of 607. Calculate and interpret a confidence interval at the 99% confidence level for the true average number of cycles to break. [Note: The article presented the results of hypothesis tests based on the t distribution; the validity of these depends on assuming normal population distributions.]A study of the ability of individuals to walk in a straight line ("Can We Really Walk Straight?" Amer. J. of Physical Anthro., 1992: 19-27) reported the accompanying data on cadence (strides per second) for a sample of n = 20 randomly selected healthy men. 95 .85 92 95 .93 .86 1.00 92 .85 78 93 93 1.05 93 1.06 1.06 96 .81 A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows: Variable N Mean Median TrMean StDev SEM cadence 20 0. 9255 0. 9300 0. 9261 0. 0809 0. Variable Min Max Q1 Q3 cadence 0 . 7800 1. 0600 0. 8525 0. 9600 a. Calculate and interpret a 95% confidence interval for population mean cadence. b. Calculate and interpret a 95% prediction interval for the cadence of a single individual randomly selected from this population. c. Calculate an interval that includes at least 99% of the cadences in the population distribution using a confidence level of 95%.Two different companies have applied to provide cable television service in a certain region. Let p denote the proportion of all potential subscribers who favor the first company over the second. Consider testing H.: p = .5 versus H,: p # .5 based on a random sample of 25 individuals. Let the test statistic X be the number in the sample who favor the first company and x represent the observed value of X. a. Describe type I and II errors in the context of this problem situation. b. Suppose that x = 6. Which values of X are at least as contradictory to H. as this one? c. What is the probability distribution of the test statistic X when H, is true? Use it to compute the P-value when x = 6. d. If Ho is to be rejected when P-values .044, compute the probability of a type II error when p = 4, again when p = .3, and also when p = .6 and p = .7. [Hint: P-value > .044 is equivalent to what inequalities involving x (see Example 8.4)?] e. Using the test procedure of (d), what would you conclude if 6 of the 25 queried favored company 1