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- 11 1 Part: 0 / 5 Part 1 of 5 X P(x) = 2 Next Part = 3 School days: The following table presents the numbers of students enrolled in grades 1 through 8 in public schools in a certain country. 1 Grade 1 2 3 Source: Statistical Abstract of the United States 4 5 6 7 2 8 Total Send data to Excel '11 Frequency (in 1000s) Consider these students to be a population. Let X be the grade of a student randomly chosen from this population. 3 = 5 S 3775 3602 3612 3587 3638 3690 3751 3790 29,445 (a) Construct the probability distribution of X. Round the answers to three decimal places. 4 6 5 7 6 00 8 7 8 9 Xx Español 000 Submit A: Ⓒ2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center |USA Today reported that approximately 25% of all state prison inmates released on parole become repeat offenders while on parole. Suppose the parole board is examining five prisoners up for parole. Let x = number of prisoners out of five on parole who become repeat offenders. X P(x) 0 0.217 1 0.378 2 0.228 3 0.150 4 0.026 5 0.001Assume that A, B, and C are subsets of a sample space S with Pr(A) = 0.75, Pr(B) = 0.55, Pr(C) = 0.25. 1. Find Pr(A'), Pr(B'), and Pr(C'). Pr(A') = Pr(B') = %3D Pr(C') = 2. If Pr(A U B) = 0.75., find Pr(A n B). Pr(A n B) = 3. Suppose that we know that B and C are disjoint events. Find Pr[B U C]. Pr[B U C] = 4. Suppose that instead of being disjoint C C B. Find Pr[B U C]. Pr[BU C] =
- NT #3 For each of the following, answer whether the distribution of X is necessarily normally distributed (YES) or not (NO), and briefly explain why. a) X = the sample mean age for random samples of size n = 50 from the population of New Yorkers of non-working age (defined as below 18 or above 64). b) X = the sample mean age for random samples of size n = 10 from the population of New Yorkers of non-working age (defined as below 18 or above 64). c) X = the sample mean age for random samples of size n = 10 from a population of individuals whose ages are known to be normally distributed.A random sample of n1 = 16 communities in western Kansas gave the following information for people under 25 years of age. x1: Rate of hay fever per 1000 population for people under 25 97 91 122 130 94 123 112 93 125 95 125 117 97 122 127 88 A random sample of n2 = 14 regions in western Kansas gave the following information for people over 50 years old. x2: Rate of hay fever per 1000 population for people over 50 94 111 100 95 110 88 110 79 115 100 89 114 85 96 What is the value of the sample test statistic? (Test the difference μ1 − μ2. Round your answer to three decimal places.)A manufacturing company employs two inspecting devices to sample a fraction of their output for quality control purposes. The first inspection monitor is able to accurately detect 99.3% of the defective items it receives, whereas the second is able to do so in 99.7% of the cases. Assume that four defective items are produced and sent out for inspection. Let X and Y denote the number of items that will be identified as defective by inspecting devices 1 and 2, respectively. Determine the following. 1) fxy(x,y) 2) fx (x)3) fY|2 4) E(x)Show all work
- A random sample of n 16 communities in western Kansas gave the following information for people under 25 years of age. X1: Rate of hay fever per 1000 population for people under 25 99 91 119 129 92 123 112 93 125 95 125 117 97 122 127 88 A random sample of ng = 14 regions in western Kansas gave the following information for people over 50 years old. Rate of hay fever per 1000 population for people over 50 94 108 100 97 111 88 11O 79 115 100 89 114 85 96 A USE SALT (1) Use a calculator to calculate x, S1, X, and sa. (Round your answers to four decimal places.) S1= x2 = (ii) Assume that the hay fever rate in each age group has an approcimately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fevert a 0.05. (a) What is the level of significance? State the null and alternate hypotheses. OHot = H2i H: 1 P2Prove the following8. The time (in minutes) that it takes a car mechanic to replace a car battery is a random variable X that follows an exponential distribution with mean 15 minutes. a) Find P(14 < X < 22). Round off your answer to two decimal places. x 22 1 P(14 < X <22) = √₂/1² 15 e 15dx = P(x < 22) — P(x < 14) x 15 122 = e 15 22 1 - e 14 or 0.8 = = 22 14 = −e 15 + e15 = 0.16254≈ 0.16 - (1 – b) Find the 80th percentile. Round off to the nearest whole number. 1 X S 15 14 e 15 t e 15 dt X 0.8 = 1 e 15 X e 15 = = 0.2 14 22 = e 15 e 15 = 0.16 x — = ln (0.2). This gives x = 24.14156 15
- Event A is my wife cooking dinner tonight. It has been determined that P(A) = 0.15. What is P(not A)?20% of Australian adults suffer from allergies. Given a random sample of ten adults, what is O (;)"Let X1,. X, be a sample from normal with mean theta and variance 1, and consider the sequence of estimators Xn = , Show, by using the definition, that Xn, is consistent. That is, show that P(|Xn-theta|<epsilon) -> 1, as n -> infinity.