02: Let X and Y have a bivariate normal distribution with parameters u, = H, = 16, af = 25 and p = 3. The value of P(-3< X < 3|Y = -4) is A. 0.9162 O A B. 0.6915 O B C. 0.6077
Q: Given µ=70.5 and o=4.6 of normal distribution, find а. р(65 80) d. p(x < 65)
A: Note: according to our guidelines we can answer only first three subparts
Q: Q2: Let X and Y have a bivariate normal distribution with parameters 4, = 2 = 1, o} = 16, a = 25 and…
A:
Q: (a) Suppose Z be a standard normal rv. Find P(Z > 1.46 or - 1.46 > Z). (b) Suppose X has a normal…
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A: Answer : The value of the X-variate is : (c) 54.
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A: Here, we have, (X, Y)~BN μx=2.9, μy=2.4, σx=2.4, σy=0.5, ρ=0.8 We know, if the joint distribution is…
Q: 4. Let X and Y have a bivariate normal distribution with respective parameters μ = 2.8,μ¸ = 110,o² =…
A: Given that, X and Y have a bivariate normal distribution. We have to find P(2<X<4/Y=
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A: q-1) given uniform distribution X~U(0,5) find P(1<x<4) = ?
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A: Solution: From the given information, X follows normal distribution with mean µ=14 and variance…
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A: Given that Total bulbs=66 Defective bulbs=13 Not Defective bulbs=53 22 bulbs are selected at random…
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A: Given that Z has standard normal distribution
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Q: Find the proportion of observation from a normal distribution N(0,1) that satisfies z>-0.52.
A: Given: the proportion of observation from a normal distribution N(0,1) that satisfies z>-0.52
Q: Q2: Let X and Y have a bivariate normal distribution with parameters 4, H2 1,of = 16, a = 25 and p=…
A: Solution
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A: Given data: Population Mean = -2 Population standard deviation = 2
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A: Population mean μ =500 Population standard deviation =20 Sample size n =25
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A: We have to find out x value for given data..
Q: Let x have a normal distribution with a mean of 91.2 and a standard deviation of 4.92. The z value…
A: Given that Mean = 91.2 , Standard deviation = 4.92 X-score = 73.98
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Q: Below is a graph of a normal distribution with mean u=-4 and standard deviation o= 2. The shaded…
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Q: If a random variable X has the gamma distribution with a = 2 and ẞ= 1, find P(1.7<X<2.3).
A: P(1.7<x<2.3)=16.24%Explanation:Given: X has a gamma distribution withα=2β=1 Probability…
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Q: Z-N(0,1) or standard normal distribution . Find P(Z at least 0.68).
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Q: If P(-X < z < +X) = 0.9198, find the value of X using Standard Normal Distribution table.
A: In this case, it is given P-X<Z<+X=0.9198.
Q: Compute: a) P(106 < Y, < 124) b) P(106 < Y, < 124 | Y, = 3.2) %3D
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A: Z be a standard normal random variable.Z~N(0,1)
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A: From the provided information, Mean (µ) = 16 Standard deviation (σ) = 2 X~N (16, 2)
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A: given data normal distributionμ = 106.4σ = 25.5n = 85
Q: Q2: Let X and Y have a bivariate normal distribution with parameters 4, = µ2 = 1,of = 16, a? = 25…
A:
Q: make a normal curve for this p= (x/n) (100) = (17/19) (100) = 89.47
A: The following information is provided in the question n = 19 p = 0.8947 We know, When X follows…
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A: Given that μ1=5.4, σ1=1.5, μ2=5.10, σ2=2, ρ=0.70 Y=6
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A: Population mean, Population standard deviation,
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- A normal distribution has a mean of µ = 100 with σ = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 90 and X = 110?Z denotes a variable having standard normal distribution. P(-0.45If X has normal distribution with mean 80 and standard deviation 15. Find the P(X > 90) a. 0.7734 b. -0.75 Oc. 0.75 O d. 0.2514The focus of a human eye is measured in diopters. For a given person, the focus is modelled by a random variable with normal distribution N(μ, o) with mean and standard deviation o. Th focus under which eye correction for nearsightedness (myopia) is considered necessay is -3 diopters. The focus of a normal eye is larger than this value. Mr Y. got arrested for going through a stop sign, not wearing glasses. Throughout the exercise the threshold is set to 5%. Based on 3 measurements on Mr. Y, we have observed an average focus of -0.94. The hypothesis testing the policeman intends to do is H₁:> -3. a. Assuming the standard deviation known and equal to o = 1.6 diopters, the rejection region of the test statistics is (-∞, -1.645) (-1.96, 1.96) * (1.645, ∞)Suppose the scores on a standardized test are normally distributed with mean μ=340μ=340 and the standard deviation σ=32σ=32. Use the normal distribution to find: The lowest test score in the top 15% bracket.Ans: The highest score in the bottom 20% bracket.Ans:In a normal distribution, a data value located 1.5 standard deviations below the mean has Standard Score: Z= In a normal distribution, a data value located 1.9 standard deviations above the mean has Standard Score: Z = In a normal distribution, the mean has Standard Score: z =Beetles' weight has a normal distribution with mean of 85g and standard deviation of 6.4g. Suppose we repeatedly took random samples of 12 of these beetles. The percentage of samples that have a sample mean weight less than 83.2g is closest to: (You should use the empirical rules to estimate this percentage) A: 2.5% B: 85% C: 100% D: 15%46. Let X have a Standard Normal Distribution. Find a such that P(X > a) = 0.80511 %3| A. – 0.66 В. — 0.76 С. — 0.86 D. – 0.96Current Attempt in Progress Let x have a normal distribution with a mean of 57.7 and a standard deviation of 4.95. The z value for x = 43.38, rounded to two decimal places, is: iLet x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with ? = 1.8%. A random sample of 10 bank stocks gave the following yields (in percents).5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1 I need help with sketching the distribution and (d) and (e)Let X follow the normal distribution with μ = 100 and σ = 10. Then P(X < 120) is closest to: a. 0.023 b. 0.046 c. 0.430 d. 0.853 e. 0.977A normal distribution has μ = 32 and σ = 5. Find the z score corresponding to x = 27. Find the z score corresponding to x = 46. Find the raw score corresponding to z = −3. Find the raw score corresponding to z = 1.1.SEE MORE QUESTIONS