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- Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with σ = 2.6%. 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 The sample mean is = 5.38%. Suppose that for the entire stock market, the mean dividend yield is μ = 4.3%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.3%? Use α = 0.01. (a) What is the level of significance? (Enter a number.)State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? H0: μ > 4.3%; H1: μ = 4.3%; right-tailedH0: μ = 4.3%; H1: μ ≠ 4.3%; two-tailed H0: μ = 4.3%; H1: μ < 4.3%; left-tailedH0: μ = 4.3%; H1: μ > 4.3%; right-tailed (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The standard normal, since we assume that x has a normal…Suppose the random variable X follows a normal distribution with mean 70 and variance 100. Then the probability is 0.6826 that X is in the symmetric interval about the mean between two numbers are O A. None O B. 70 and 90 O C. -1 and 1 OD. 60 and 80Given that z is a standard normal random variable, compute P(z < -2.14). Select one: a. 0.0154 O b. 0.9834 C. 0.9842 d. 0.9838 e. 0.0166 of. 0.0158 g. 0.0162 O h. 0.9846
- Suppose a random variable X is best described by a normal distribution with u=60 and o2=144 Find the z-score that corresponds to the value X = 84. O 1. 12 O 2. 1.2 O 3. 2 O 4. None of theseFor a random variable that is normally distributed, with µ = 1300 and σ = 250, determine the probability that a simple random sample of 9 items will have a mean that is greater than 1500 b. between 1425 and 1500 c. less than 225The random variable X is related to e2. Find P(X>0.4).
- Let X be the random variable with the following distribution: x: 1 2 3 4 5 P(X=x): k 0.04 0.3 0.2 0.3 Find the mean four decimal placesSuppose that random variable X follows a normal distribution with mean 20 and standard deviation 10. Let Y=10-2X. The probability P(5sY<15) is closest to O 0.2417 0.4995 0.4332 O 0.0279 O 0.1915Suppose that X has a Weibull distribution with β = 2 and δ = 2400. Determine the following. a. P(X > 5000) = b. For an exponential random variable with the same mean as the Weibull distribution P(X > 5000) =
- Suppose X is a normal random variable with mean u = 17.5 and o = 6. A random sample of size n = 24 is selected from this population. a. Find the distribution of b. Find P(X < 14) and P(X < 14) Find P(15Suppose x is a normally distributed random variable with mean u =0 and standard deviation o = 1. Find the probability that x < 0.64. O 73.89% 76.11% 23.89% O 26.11%Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with o = 2.4%. 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 The sample mean is x = 5.38%. Suppose that for the entire stock market, the mean dividend yield is u = 4.8%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.8%? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? Ho: H = 4.8%; H1: µ ± 4.8%; two-tailed Ho: H = 4.8%; H1: µ > 4.8%; right-tailed Ho: H = 4.8%; H1: µ 4.8%; H1: µ = 4.8%; right-tailed (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The Student's t, since n is large with unknown o. The Student's t, since we assume that x has a normal distribution with known o. The standard normal, since…SEE MORE QUESTIONS
