If the independent r.v.'s X and Y are distributed as N(µ₁, 0²) and N(µ2, 02), respectively, then show that the pair (X,Y) has the bivariate normal distribution, and specify the parameters involved.
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- An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in × = 18.17 kgf/cm² for the modified mortar 42 be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. = 0 versus H₂: M₁ M₂ > 0 at level 0.01. (m = 42) and y = 16.87 kgf/cm² for the unmodified mortar (n = 30). Let and M1 (a) Assuming that 0₁ = 1.6 and 0₂ 1.3, test Ho: M1 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) 42 Z = P-value = State the conclusion in the problem context. Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0. O Fail to reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds…Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than −1.11 and draw a sketch of the region?Draw the standard normal curve that corresponds to an area between z = 0 and z = 2. Then, use the standard normal table to find the area under the curve between these two z-scores.
- Let X and Y have a bivariate normal distribution with μ = 70, X Find (a) E(Y|X= 76) (b) Var(Y | X= 76) (c) P (Y≤ 861X=76) 0²: X = 100, = 80, My o² = 169, and p=5/13.Find a z-score satisfying the given condition. For any normal distribution, what is the value of P(xsp) and P(x2u)? P(xsp) =D and P(x2 u) =D (Type integers or decimals.)Find the Z score such that 43.9% of the area under the standard normal curve is below that score.
- Find the z-score such as the standard normal curve to the left is 0.49Suppose the flow of current (milliamps) in wire strips of a certain type under specified conditions can be modeled with a normal distribution having mu (mean) = 20 and sigma (SD) = 1 (hint: think about how the corresponding density curve relates to the standard normal curve). What proportion of strips will have of between 18.5 and 22 milliamps?Find the area under the standard normal curve to the right of z = 2.04 z= 1.09 Genetics: Pea plants contain two genes for seed color, each of which may be Y (for yellow seeds) or G (for green seeds). Plants that contain one of each type of gene are called heterozygous. According to the Mendelian theory of genetics, if two heterozygous plants are crossed, each of their offspring will have probability 0.75 of having yellow seeds and probability 0.25 of having green seeds. One hundred such offspring are produced. Approximate the probability that more than 30 have green seeds.
- About % of the area under the curve of the standard normal distribution is between z=−0.153z=-0.153 and z=0.153z=0.153 (or within 0.153 standard deviations of the mean).The pressure of air (in MPa) entering a compressor is measured to be X = 8.5 ± 0.2, and the pressure of the air leaving the compressor is measured to be Y = 21.2 ± 0.3. The intermediate pressure is therefore measured to be P = √XY = 13.42 . Assume that X and Y come from normal populations and are unbiased. a) From what distributions is it appropriate to simulate values X* and Y*? b) Generate simulated samples of values X*, Y*, and P*. c) Use the simulated sample to estimate the standard deviation of P. d) Construct a normal probability plot for the values P*. Is it reasonable to assume that P is approximately normally distributed? e) If appropriate, use the normal curve to find a 95% confidence interval for the intermediate pressure.Consider a test of Ho : u = 4. In each of the following cases, give the rejection region, assuming that the test statistics follows a standard Normal distribution. (a) H. : u > 4, a = 0.1 (b) H. : u <4, a = 0.01 (c) Ha : 4# 4, a = 0.05