If X is a normal variate with mean 18 and variance 625, find P(-31 18 ). 1.96 1 [ Given L dt = 0.9750021 } 12 %3D e
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- Suppose that the weight of an newborn fawn is Uniformly distributed between 2.2 and 3 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. The probability that a newborn fawn will be weigh more than 2.36 is P(x > 2.36) = P(x > 2.6 | x < 2.9) = Find the 8th percentile.Suppose Z is distributed as a Standard Normal variable. Calculate Pr(-1.54 < Z < -0.13) How do I calculate problems like this using Excel & without excel?The confidence interval for the population variance was found to be 86.233 to 456.955. However, a confidence interval for the population standard deviation is needed. Recall that the standard deviation is the positive square root of the variance. Therefore, the confidence interval for the standard deviation can be found by taking the positive square root of the lower and upper bounds for the variance. Substitute o? = 86.233 to find the lower bound for the confidence interval of the population standard deviation, rounding the result to two decimal places. lower bound of o - V lower bound of o? Substitute o2 = 456.955 to find the upper bound for the confidence interval of the population standard deviation, rounding the result to two decimal places. upper bound of a = V upper bound of a Therefore, a 95% confidence interval for the population standard deviation for the underpricing-overpricing indicator is from a lower bound of to an upper bound of
- I am having difficulty with The Analysis of Biological Data chapter 10 question 15AP, I need to check my solutions but there are none posted for me to review. If I could compare answers that would be great.A random sample of 28 lunch orders at Noodles & Company showed a standard deviation of $6.24. Find the 99 percent confidence interval for the population standard deviation. Use Excel to obtain χ2L=CHISQ.INV(α/2,d.f.)χL2=CHISQ.INV(α/2,d.f.) and χ2U=CHISQ.INV.RT(α/2,d.f.)χU2=CHISQ.INV.RT(α/2,d.f.) (Round your answers to 2 decimal places.) The 99% confidence interval is from _________ to____________ .A sample of n = 64 scores is selected from a population with µ = 80 and with σ = 24. On average, how much error is expected between the sample mean and the population mean?
- You believe both populations are normally distributed, but you do not know the standard deviations for either. And you have no reason to believe the variances of the two populations are equal You obtan a sample of size nį = 25 with a mean of 1 population. You obtain a sample of size n2 s2 = 13.9 from the second population. 63.7 and a standard deviation of s = 13.2 from the first 19 with a mean of g 67 and a standard deviation ofConsider regression model Y; = R + B,X; + B;X +U,with hypothetical data Ev = 50, Ex = 20. Eyx = 60, E = 30, Σχ-0.5 Σν-300 Find OLS estimates of B, and B and derive their variances. Calculate R'and interpret it Construct 95% confidence interval for B, and interpret it Q1: n= 25 i. ii. 111In a trivariate distribution the simple coefficient of correlation are as follows if r12 = 0.76, rı3= 0.75 and r²3= 0.71 calculate the coefficient of partial correlation r 12.3
- Answer allA random sample of 10 observations from population A has sample mean of 152.3 and a sample standard deviation of 1.83. Another random sample of 8 observations from population B has a sample standard deviation of 1.94. Assuming equal variances in those two populations, a 99% confidence interval for μA − μB is (-0.19, 4.99), where μA is the mean in population A and μB is the mean in population B. (a) What is the sample mean of the observations from population B? (b) If we test H0 : μA ≤ μB against Ha : μA > μB, using α = 0.02, what is your conclusion?let Y=5X+10 and X be normally distributed with a mean 10 and varience 25. find P(Y<54).