Hello expert, I need to solve the question step by step and in clear writing If he X~N(μ, σ²) 1) The mean deviation in this distribution is 2) Q1 0.670, Q3 ==μ + 0.67σ σ
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- Respiratory Rate Researchers have found that the 95 th percentile the value at which 95% of the data are at or below for respiratory rates in breath per minute during the first 3 years of infancy are given by y=101.82411-0.0125995x+0.00013401x2 for awake infants and y=101.72858-0.0139928x+0.00017646x2 for sleeping infants, where x is the age in months. Source: Pediatrics. a. What is the domain for each function? b. For each respiratory rate, is the rate decreasing or increasing over the first 3 years of life? Hint: Is the graph of the quadratic in the exponent opening upward or downward? Where is the vertex? c. Verify your answer to part b using a graphing calculator. d. For a 1- year-old infant in the 95 th percentile, how much higher is the walking respiratory rate then the sleeping respiratory rate? e. f.Use the table of values you made in part 4 of the example to find the limiting value of the average rate of change in velocity.Find the average rates of change of f(x)=x2+2x (a) from x1=3 to x2=2 and (b) from x1=2 to x2=0.
- The deviation of each observation (x) from the mean of the predictor variable is positive for every point in what quadrants? * Y (2) 1) 3 4)I need help with these two questions. Here are the data table if needed. The absolute viscosity for sunflower oil is supposed to average 0.0311 Pa⋅s at 38 ∘C. Suppose a food scientist collects a random sample of 4 quantities of sunflower oil and computes the mean viscosity for his sample to be x¯=0.0318 Pa⋅s at 38 ∘C. Assume that measurement errors are normally distributed and that the population standard deviation of sunflower oil viscosity is known to be σ=0.0009 Pa⋅s. The scientist will use a one‑sample z‑test for a mean, at a significance level of α=0.01, to evaluate the null hypothesis, H0:μ=0.0311 Pa⋅s against the alternative hypothesis, H1:μ≠0.0311 Pa⋅s. Complete the scientist's analysis by calculating the value of the one-sample z‑statistic, the p‑value, and then deciding whether to reject the null hypothesis. First, compute the z‑statistic, z. Provide your answer precise to two decimal places. Avoid rounding within calculations. z= Determine the…Approximate the population mean and standard deviation of age for males. By the way, to find the midpoints we are adding vertically and then dividing by 2 (ex: 0+10/2=5, 10+20/2=15).
- Please do not give solution in image format thanku The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 70, and the standard deviation is 2. You wish to test H0 : μ =70 versus H1 : μ ≠70 with a sample n = 5 specimens. Find β for the case where the true mean heat evolved is 73. Use the α that you calculated in question 6. Report to three digits after decimal point Problem #6: α= 0.907#19 solve using applications of normal distributionA population of values has a normal distribution with mean of 159.4 and standard deviation of 80.5. You sample 217 values from the population. a) Get the z-score for a sample mean of 144.1. For this problem you use the z score formula z=(¯x−μ)√nσz=(x¯-μ)nσ b) This z-score tells you how many the sample mean of 144.1 is above or below the population mean μμ . c) Find the probability that a sample mean is greater than 144.1. Enter your numerical answers as numbers accurate to at least 4 decimal places.
- Based on the statement “Can you conclude that the mean wind speed when it is less than 12m/s? Use α = 0.10.”, it is a _______. I. One tail test on the right-hand side II. One tail test on the left-hand side III. Two tails test on both right-and-left hand sidesYou have information to suggest that a certain continuous variable of a population has a mean of μ = 9.13 and a standard deviation of o = 6.67. You are to randomly pick n = 71 individuals from this population and observe the value of the population variable on each. This value is to measured as X₁. After the random sample of n = 71 has been taken, you are asked to consider the behavior of the statistic X. (a) Complete the statement below. Use as many decimals as you can. The distribution of X is approximately Normal with a mean x = 9.13 and a standard deviation x = 0.7916 (b) Find the probability that mean of the sample of n = 71 is between 8.53 and 9.53. Use at least four decimals in your z-values, and enter your answer to at least four decimals.a b Absolute standard deviation = Coefficient of variation = y = 1.20 (+0.02) x 10-8-3.60(±0.2) × 10-⁹ x Result = Absolute standard deviation = Coefficient of variation = C Result = d y=90.95 (±0.08) - 89.40(±0.06) +0.200(+0.004) Absolute standard deviation == Coefficient of variation = Result = ± y= y = 0.0040 (+0.0005) x 10.28 (+0.02) × 395(+1) Result = ± Absolute standard deviation = Coefficient of variation = % 329(+0.03) x 10-14 1.47 (+0.04) x 10-16 % % H % ±