Construct a Binomial probability distrubtion table using n = record the mean and standard deviation of the distribution below. Where necessary, round all numbers to four decimal places. 5 and p = 0.21. Next, P(x) O =
Q: The mean percent of childhood asthma prevalence in 43 cities is 2.27%. A random sample of 33 of…
A: Given : μ = 2.27 , σ= 1.27 and n= 33 By Central Limit Theorem, x̅ follows a Normal distribution…
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Q: find the mean of this probability distribution, round your answer to one decimal place. X 0,1,2,3…
A: x P(X) x*P(x) 0 0.3 0 1 0.2 0.2 2 0.15 0.3 3 0.35 1.05 ∑x*P(x) = 1.55
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Q: Tony runs a cell phone kiosk in the mall and has 244 potential customers each day. He has a 4 %…
A: Given,n=244p=0.04mean(μ)=npμ=244×0.04=9.76Standard deviation (σ)=np(1-p)σ=244×0.04×0.96=3.06098
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A: Givenn=235p=0.06using normal approximationMean(μ)=np=235×0.06=14.1standard…
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Q: Assume that adults have IQ scores that are normally distributed with a mean of μ=105 and a standard…
A: Let "X" be the IQ score of the adults.
Q: About _____% of samples of 33 cities will have a mean childhood asthma prevalence greater than…
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Q: The mean percent of childhood asthma prevalence in 43 cities is 2.38%. A random sample of 32 of…
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Q: P(x) 0.25 1 0.15 2 0.05 0.55 Find the mean of this probability distribution. Round your answer to…
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Q: Write the normal probability for the shaded region of the graph and find its value. Select the…
A: Given,n = 16p = 0.55q = 0.45
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A: Given data: Pedestrian Intoxicated Driver Intoxicated Yes No Total Yes 56 84 140 No 215…
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A: given;X has binomial distributed.i.e. X~binomial(n,p)where n = 28, p=0.28
Q: The mean percent of childhood asthma prevalence in 43 cities is 2.21%. A random sample of 32 of…
A: Guven:Mean (µ) = 2.21Standard deviation (σ) = 1.36Sample size (n) = 32
Q: The mean percent of childhood asthma prevalence in 43 cities is 2.41%. A random sample of 31 of…
A: Guven:Mean (µ) = 2.41Standard deviation (σ) = 1.36Sample size (n) = 31
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A: Solution: The given table of data is
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A: Given that Mean percent of childhood ashthma prevalence =2.42% =0.0242 Sample…
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A: We are going to find the value by the help of formula in binomial distribution
Q: The mean percent of childhood asthma prevalence in 43 cities is 2.23%. A random sample of 34 of…
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Q: Find the mean, variance, and standard deviation for each of the values n of p and when the…
A: Conditions for binomial distribution:1. There are two mutually exclusive outcomes, success or…
Q: The table summarizes results from 990 pedestrian deaths that were caused by automobile accidents.…
A: Total number of pedestrian deaths is n = 990.
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A: Note: Hi there! Thank you for posting the question. As there are multiple sub parts, according to…
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- Juan makes a measurement in a chemistry laboratory and records the result in his lab report. The standard deviation of students lab measurements is o= 8 milligrams. Juan repeats the measurement 2 times and records the mean x of his 2 measurements. What is the standard deviation of Juan's mean result?X P(x) 0.2 1 0.3 2 0.05 3 0.45 Find the mean of this probability distribution. Round your answer to one decimal place.Find the indicated probability and interpret the result. From 1975 through 2020, the mean annual gain of the Dow Jones Industrial Average was 653. A random sample of 34 years is selected from this population. What is the probability that the mean gain for the sample was between 400 and 800? Assume o = 1540. The probability is (Round to four decimal places as needed.) Interpret the result. Select the correct choice and fill in the answer box to complete your choice. (Round to two decimal places as needed.) O A. About OB. About OC. About OD. About % of samples of 46 years will have an annual mean gain between 400 and 800. % of samples of 34 years will have an annual mean gain between 653 and 800. % of samples of 34 years will have an annual mean gain between 400 and 800. % of sampl of 34 years will have an annual mean gain between 400 and 653.
- Instructions: Use the Standard Normal Table to find the following probabilities. P(z -1.37) = P(z 0.04) = P(-1.37 < z < 1.68) =Thirty percent of people did not visit their doctor's, offices last year. Let x be the number of adults in a random sample of 12 adults who did not visit their doctors’ offices last year. The standard deviation of the probability distribution of x is approximately Could you show the steps and equations used, please?The table summarizes results from 990 pedestrian deaths that were caused by automobile accidents. DriverIntoxicated? Pedestrian Intoxicated? Yes No Yes 47 86 No 237 620 If one of the pedestrian deaths is randomly selected, find the probability that the pedestrian was intoxicated or the driver was not intoxicated.Report the answer as a percent rounded to one decimal place accuracy. You need not enter the "%" symbol.prob = % Add Work
- The mean percent of childhood asthma prevalence in 43 cities is 2.38%. A random sample of 30 of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8%? Interpret this probability. Assume that o = 1.25%. The probability is (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as needed.) O A. About % of samples of 43 cities will have a mean childhood asthma prevalence greater than 2.8%. O B. About % of samples of 30 cities will have a mean childhood asthma prevalence greater than 2.38%. O C. About % of samples of 30 cities will have a mean childhood asthma prevalence greater than 2.8%. Click to select your answer(s). P Type here to search 99+ a (? 1:42 PM 4/11/2021 ho insen 144 ese & t backspace %23 3. %24 4 6. 8 2. Y W tab H. J. K pase 5 RWrite the binomial probability and the normal probability for the shaded region of the graph. Find the value of each probability and compare the results. AP(x) 0.24- n = 16 p= 0.5 0.2- 0.16- 0.12- 0.08- 0.04- 0- 2 4 6 8 10 12 14 16 Write the binomial probability for the shaded region of the graph and find its value. Select the correct choice below and fill in the answer box within your choice. (Round to four decimal places as needed.) O A. P(4Assume that adults have IQ scores that are normally distributed with a mean of u = 105 and a standard deviation a = 15. Find the probability that a randomly selected adult has an IQ between 91 and 119. Click to view page 1 of the table. Click to view page 2 of the table. The probability that a randomly selected adult has an IQ between 91 and 119 is. (Type an integer or decimal rounded to four decimal places as needed.) Enter your answer in the answer box. MacBook Air esc 888 %23 & 2 3 4 5 6 7 8 Q E R T Y tab P A D G H caps lock K V в N M > ? control option command command option