4. A proof-reader discovered 200 misprints in a book containing 800 pages. Assuming that the misprints occur at random, find the probability that a particular page contains a) no misprints b) one or two misprints c) more than two misprints
Q: normal distribution
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- = According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0 10. A random sample of 1,200 Americans results in 132 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. O A. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the sample size n is more than 5% of the population OB. This is not necessarily evidence that the proportion of Americans who are afraid to fly has increased above 0.10 because the value of np(1-p) is less than 10. OC. This is not necessarily evidence that the proportion of Americans who are afraid to fly has…All parts pleasetopic: function of random variables
- In the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2, 36. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. If the number of the slot the ball falls into matches the number you selected, you < win $35; otherwise you lose $1. Complete parts (a) through (g) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) Construct a probability distribution for the random variable X, the winnings of each spin. P(x) X 35 - 1 (Type integers or decimals rounded to four decimal places as needed.) OODModeling and Simulation (Poisson Distribution) Q1) The number of hurricanes hitting the coast of Florida annually has a Poisson distribution with a mean of 0.7. What is the probability that more than one hurricanes will hit the Florida coast in a year ?Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test Ho: p=0.5 versus H₁: p > 0.5 n = 250; x = 145; a = 0.1 Click here to view page 1 of the table Click here to view page 2 of the table Calculate the test statistic, Zo Zo (Round to two decimal places as needed.) = DE COOK
- Normal distribution of a quantitative trait. (a) The distribution of heights in 82 college students. (b) A frequency distribution for the heights of students shown in (a). Is height a discontinuous (discrete) trait, or does it follow a continuum?normal distributionNormal Distribution] The average return on large-cap stocks is 10%. The returns on large-cap stocks is normally distributed with standard deviation of 5% What is the probability that a large-cap stock has a return of less than 5%?
- Maximum Likelihood Estimator (MLE) Poisson Distribution (a) A call center receives calls with a poisson distribution with an average of calls per hour. Let X₁, X2,... Xn be a random sample, then find the MLE of of the Poisson distribution. (b) Suppose we observe 10, 15, 25, 10, 30 for 5 randomly selected hours, what is the MLEFind indicated probability using standard normal distribution. Standard normal table. properly labeled accurate sketch of region being described. Correct region is shaded. Sketches may be added before or after converting to z values easier to convert z first they locate values on number lineSubgroup of Participants Supine SBP Mean (SD) Standing SBP Mean (SD) Percent with > 5.0 Decrease Percent with > 5.0 Increase White women (N = 301) 120.4 (18.4) 120.0 (18.8) 32.6% 32.2% African-American women (N = 107) 122.7 (19.3) 122.8 (19.4) 30.2% 31.9% White men (N = 339) 121.2 (18.9) 120.2 (19.9) 39.5% 25.6% African-American men (N = 95) 124.7 (20.4) 125.9 (21.6) 13.7% 51.3% Total (N = 842) 121.5 (19.0) 121.1 (19.5) 29.8% 31.7% Would it be possible to analyze data in table 2 with a chi-square test? if yes, specify how