Instructions: Use the Standard Normal Table to find the following probabilities. P(z < -2.45) = P(z > -1.37) = P(z < 1.99) = P(z > 0.04) = | P(-1.37 < z < 1.68) =
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A: It is given that the standard normal random variable( Z ) is equal to -0.68.
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A: a) P(-1.24<z<0)=?=P(Z<0)-P(Z<-1.24)=0.5000-0.1075=0.3925 z-table:
Q: 1 What is the probability that a standard normal variable will be equal to -0.68? 0 1 .7517 5120…
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- hwm7 7 Use the following probabilities to answer the question. P(A) =0.6, P(B)=0.35 AND P(AandB)=0.05. P(notB notA)= _%Many track hurdlers believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The data lists the number of wins for track hurdlers in the different starting positions. Calculate the chi-square test statistic x to test the claim that the probabilities of winning are the same in the different positions. Use α = 0.05. The results are based on 240 wins. Starting Position 3 4 5 6 1 2 Number of Wins 44 32 33 50 36 45 O A. 6.750 O B. 12.592 O C. 15.541 O D. 9.326 CUsing historical records, the personnel manager of a plant has determined the probability of X, the number of employees absent per day. It is 2 4 6 7 P(X) 0.005 0.0246 0.31 0.33910.2195 0.080.01850.0033 4. Find the following probabilities. Be sure to give your answer to 4 decimal places. A. P(2 5) Probability = С. Р(Х < 4) Probability: %3D 出
- A manufacturer of plumbing fixtures has developed a new type of washerless faucet. Let p = P(a randomly selected faucet of this type will develop a leak within 2 years under normal use). The manufacturer has decided to proceed with production unless it can be determined that p is too large; the borderline acceptable value of p is specified as 0.10. The manufacturer decides to subject n of these faucets to accelerated testing (approximating 2 years of normal use). With X = the number among the n faucets that leak before the test concludes, production will commence unless the observed X is too large. It is decided that if p = 0.10, the probability of not proceeding should be at most 0.10, whereas if p = 0.30 the probability of proceeding should be at most 0.10. (Assume the rejection region takes the form reject H, if X 2 c for some c. Round your answers to three decimal places.) What are the error probabilities for n = 10? n USE SALT P-value = 0.07 B(0.3) =Suppose that the speed at which cars go on the freeway is normally distributed with mean 80 mph and standard deviation 7 miles per hour. Let X be the speed for a randomly selected car. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X - N( 80 | 7 b. If one car is randomly chosen, find the probability that it is traveling more than 79 mph. c. If one of the cars is randomly chosen, find the probability that it is traveling between 82 and 87 mph. d. 71% of all cars travel at least how fast on the freeway? mph. Hint:Q4. Assume that daily evaporation rates (E) have a uniform distribution with a = 0 and b = 0.35 in/day. Determine the following probabilities: (a) P(E > 0.1); (b) P(E < 0.22); and (c) P(E = 0.2).
- The accompanying tree diagram represents a two-stage experiment. (Let x = 0.3, y = 0.7, r = 0.7, s = 0.3, t = 0.4, and w = 0.6.) Use the diagram to find the following probabilities. (Round your answers to three decimal places.) (a) Find P(A) · P(D | A).(b) Find P(B) · P(D | B).(c) Find P(A | D).Find the following probabilities; P(0<z<2.32)Suppose that the weights of people who work in an office building are normally distributed with a mean of u = 165 Ib. and a standard deviation of o = 25 lb. What is the chance that the total weight of 5 people is more than 1000 Ibs., i.e., what is P(X1+ ... + X5 > 1000)? [Hint: try to express this as an X-style problem.
- Find the Z score associated with each of the following tail probabilities:Which of the following expressions represents the number of distinguishable permutations of the letters of the word CONCLUSIONS? 11! 8! C. 11! 2! 2! 2! 2! 11! А. 11! В. С. D. 2! 2! 2!A survey showed the following probabilities of having X number of dogs per househol X = number of dogs P(X) 1 2 3 4 5 0.3 0.25 0.15 0.15 0.1 0.05 what is the mean ? what is the standard deviation ?