The CDF of a random variable Y that is discrete is stated by the image attached below. Find the PMF based on the given CDF
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The CDF of a random variable Y that is discrete is stated by the image attached below. Find the PMF based on the given CDF.
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- A QR code photographed in poor lighting, so that it can be difficult to distinguish black and white pixels. The gray color (X) in each pixel is therefore coded on a scale from 0 (white) to 100 (black). The true pixel value (without shadow) the code is Y = 0 for white, and Y = 1 for black. We treat X and Y as random variables. For the highlighted pixel in the figure is the gray color X = 20 and the true pixel value is white, i.e. Y = 0. We assume that QR codes are designed so that, on average, there are as many white as black pixels, which means that pY (0) = pY (1) = 1/2. In this situation, X is continuously distributed (0 ≤ X ≤ 100) and Y is discretely distributed, but we can still think about the simultaneous distribution of X and Y. We start by defining the conditional density of X, given the value of Y : fX|Y(x|0) = "Pixel is really white" fX|Y(x|1) =" Pixel is really balck " Use Bayes formula as given in the picture and find the probability for x = 20 like in the picture.A study was conducted to examine the difference between the average cholesterol levels in the blood of men and females. It can be assumed that cholesterol levels follow a normal distribution in both populations and that the distribution in the populations is similar (e. Assume that cholesterol levels follow a normal distribution in both populations with similar variances). An equal number of measurements were made on men and women. The results were entered into R and the R code and output can be seen below: t.test(kol$mael-kol$kyn,var.equal = T) %3D Two Sample t-test data: kol$mael by kol$kyn t = 5.7731, df = 18, p-value alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 15.28775 32.78050 = 0.000018 sample estimates: mean in group Men mean in group women 219.2115 195.1774 1. Which statement is best based on the results of the study? Use a=0.05. a. The P-value is lower than a and therefore can be rejected HO b. The P-value is higher than a…Please answer q's a, b, and c For this case, a=3, b=9, and label the random variable A. Student ID number: 300546514
- Please answer q's d, e (i) and (ii) For this case, a=3, b=9, and label the random variable A. Student ID number: 300546514Let a be a random variable representing the percentage of protein content for early bloom alfalfa hay. The average percentage protein content of such early bloom alfalfa should be u = 17.2%. A farmer's co-op is thinking of buying a large amount of baled hay but suspects that the hay is from a later summer cutting with lower protein content. A small amount of hay was removed from each bale of a random sample of 50 bales. The average protein content from the samples was determined by a local agricultural college to be -15.8% with a sample standard deviation of S = 5.3%- At a = .05, does this hay have lower average protein content than the early bloom alfalfa?In the airline business, "on-time" flight arrival is important for connecting flights and general customer satisfaction. Is there a difference between summer and winter average on-time flight arrivals? Let x1 be a random variable that represents percentage of on-time arrivals at major airports in the summer. Let x2 be a random variable that represents percentage of on-time arrivals at major airports in the winter. A random sample of n1 = 16 major airports showed that x1 = 74.9%, with s1 = 5.3%. A random sample of n2 = 18 major airports showed that x2 = 70.2%, with s2 = 8.6%. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value a small amount and thereby produce a slightly more "conservative" answer. (a) Does this information indicate a difference (either way) in the population mean percentage of on-time arrivals for summer compared to winter? Use α = 0.05. (i) What is the…
- Burger Queen is conducting a study at its Linstead branch. Data of number of cars that orders at its drive through between the hours 8pm and 10pm on Sundays was obtained and shown in the table below. Number of Cars Frequency 4 4 5 5 6 10 7 15 8 20 From the table of random numbers attached/below, use the column of numbers starting with 99 to answer the following: Number of Cars Frequency 4 4 5 5 6 10 7 15 8 20 a) Simulate the number of cars for 14 Sundays between 8pm and 10 pm at this Burger Queen drive through.I have uploaded my question as image . please give answerA study of fox rabies in a country gave the following information about different regions and the occurrence of rabies in each region. A random sample of n1 = 16 locations in region I gave the following information about the number of cases of fox rabies near that location. x1: Region I Data 1 9 9 9 7 8 8 1 3 3 3 2 5 1 4 6 A second random sample of n2 = 15 locations in region II gave the following information about the number of cases of fox rabies near that location. x2: Region II Data 1 1 5 1 6 8 5 4 4 4 2 2 5 6 9 What is the value of the sample test statistic? (Test the difference μ1 − μ2. Do not use rounded values. Round your final answer to three decimal places.)
- A sporting goods manufacturing company wanted to compare the distance traveled by golf balls produced using four different designs. Nine balls were manufactured with each design and were brought to the local golf course for the club professional to test. The order in which the balls were hit with the same club was randomized to the pro did not know which type of ball was being hit. All 36 balls were hit (assuming the environmental conditions played no role). The resulting distances in yards were recorded and are shown in the following table: Design 1 Design 2 Design 3 Design 4 206.32 217.08 226.77 230.55 207.94 221.43 224.79 227.95 206.19 218.04 229.75 231.84 204.45 224.13 228.51 224.87 209.65 211.82 221.44 229.49 203.81 213.90 223.85 231.10 206.75 221.28 223.97 221.53 205.68 229.43 234.30 235.45 204.49 213.54 219.50 228.35 Write a set of hypotheses to test the mean…A random selection of volunteers at a research institute have been exposed to a typical cold virus. After they started to have cold symptoms, 15 of them were given multivitamin tablets formulated to fight cold symptoms. The remaining 15 volunteers were given placebo tablets. For each individual, the length of time taken to recover from the cold is recorded. At the end of the experiment the following data are obtained. Days to recover from a cold Treated with multivitamin Treated with placebo 3.0, 5.6, 1.5, 6.8, 3.8, 7.5, 5.8, 4.6, 2.4, 5.0, 7.5, 5.0, 2.6, 1.7, 6.7 4.9, 6.1, 4.9, 4.2, 3.4, 5.5, 5.6, 3.4, 7.9, 6.8, 4.8, 4.2, 5.7, 2.2, 4.0 Send data to Excel Send data to calculator It is known that the population standard deviation of recovery time from a cold is 1.8 days when treated with multivitamin tablets, and the population standard deviation of recovery time from a cold is 1.5 days when treated with placebo tablets. It is also known that both populations are approximately normally…Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The mean shopping time, μ , spent by customers at the supermarkets has been reported to be 36 minutes, but executives hire a statistical consultant and ask her to determine whether it can be concluded that μ is greater than 36 minutes. The consultant plans to do a statistical test and collects a random sample of 50 shopping times at the supermarkets. Suppose that the population of shopping times at the supermarkets has a standard deviation of 16 minutes and that the consultant performs her hypothesis test using the 0.05 level of significance.
