Given the normally distributed random variable X with σ = 5 and P(X ≥ 25)=.0526, find µ.
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A: μ=0, σ2=1
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A: Solution : = 145 = 150 s =13.6 n = 40
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A: From the given information, Consider, Sample size (n) = 14 Population with μ=67 and σ=19. Thus,
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A: We have given that n1=30 ,n2= 40 ,xbar=56,xbar2=51,s1=8.2 ,s2=6.9
Q: Let x be a random variable that represents blood glucose level after a 12-hour fast. Let y be a…
A: The provided information are:
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- Find the corresponding z-score for a sample of n = 16 and M = 38 taken from a population with a mean of μ= 40 and σ = 8.Assume the random variable X is normally distributed with a mean of μ=50μ=50and a standard deviation of σ=7.σ=7.Compute and find the probability P(X>35).For a population with μ = 80 and σ = 10,the z-score corresponding to X = 85 would be?
- Help me pls3. Let X1, .… , Xn be a random sample. Find the mle of 0 for the cases that the population pdf/pmf is, f(r;:0) = 0, %3D otherwiseA psychologist developed a standardized test for selecting “gifted” children. In the population, the test scores are normally distributed, with μ=75μ=75 and σ=10σ=10. We randomly sampled 49 gifted students and found out the mean test score of these 49 students is 77. Over repeated samples (with a fixed n=49n=49), what is the probability of getting a sample mean that is equal to or greater than the one we got? A. 0.42074 B. 0.91924 C. 0 D. 0.57925 E. 0.08076
- a sample is selected from a normal population with u=40 and σ =12. If the sample mean of M=36 corresponds to z=-2.00, then the sample size is n=64. is it true or falseA random sample of size 36 from a population with known variance, o = 9, yields a sample mean of x = 17. Find ß, for testing the hypothesis H,: u = 15 versus H1 : =16. Assume a 0.05. %3D %3DX is a normally distributed random variable with unknown µ and an unknown variance, o. X and s are the unbiased, sample mean and sample variance estimates of u and o obtained from a small sample size below 25 (n<25). The 95% C.I. is calculated as: Q9. s2 D. µ ± tn-1. D. µt tn-1 s2 A. X± 1.96 B. μ+ 1.96, C. X±tn-1, n n n О А. А В. В ОС. С O D. D
- A biologist measured the lengths of 1000 cuckoo bird eggs. The results are summarized below: Length (in millimeters) 18.75- 19.75 19.75 - 20.75 20.75 -21.75 21.75 - 22.75 22.75- 23.75 23.75 - 24.75 24.75- 25.75 Percent of Eggs 0.8 4.0 17.3 37.9 28.5 10.7 0.8 a) What percent of the group of eggs was less than 21.75 mm long? b) What is the probability that one of the eggs selected at random was at least 20.75 mm long but less than 24.75 mm long?Let x be a random variable that represents blood glucose level after a 12-hour fast. Let y be a random variable representing blood glucose level 1 hour after drinking sugar water (after the 12-hour fast). Units are in milligrams per 10 milliliters (mg/10 ml). A random sample of eight adults gave the following information. Σx = 63.8; Σx2 = 520.96; Σy = 89.8; Σy2 = 1046.96; Σxy = 731.35 x 6.2 8.7 7.0 7.5 8.1 6.9 10.0 9.4 y 9.7 10.8 10.5 11.3 14.2 7.0 14.1 12.2 (c) Find the sample correlation coefficient r and the sample coefficient of determination r2. (Round your answers to three decimal places.) r = r2 = (d) If x = 7.0, use the least-squares line to predict y. (Round your answer to two decimal places.) y = Find an 80% confidence interval for your prediction. (Round your answers to two decimal places.) lower limit mg/10 ml upper limit mg/10 ml (e) Use level of significance 1% and test the claim that the population correlation coefficient ρ…