A random sample of size n1 = 14 is selected from a normal population with a mean of 76 and a standard deviation of 9. A second random sample of size n2 - 7 is taken from another normal population with mean 71 and standard deviation 16. Let X1 and X2 be the two sample means. Find: (a) The probability that X – X2 exceeds 4. (b) The probability that 4.3 < X1 – X2 < 5.1. Round your answers to two decimal places (e.g. 98.76). (a) i (b) i
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- Let be a normally-distributed random variable with mean u= 100 and variance 2 = 16: Select which statement is true. P(94 < X< 106) = 0.68 approximately. P(92 < X< 108) = 0.68 approximately. P(86 < X<114) = 0.68 approximately. none of these P(96 < X< 104) = 0.68 approximately.Assume that adults have IQ scores that are normally distributed with a mean of mμ=100 and a standard deviation σ= 15. Find the probability that a randomly selected adult has an IQ less than 127. The probability that a randomly selected adult has an IQ less tha 127 isa) Find such that P( X < ? ) = 0.9332, where X is a normal random variable with mean = 10 and standard deviation = 2.5. b) Find such that P( X > ?) = 0.1230, where X is a normal random variable with mean = 10 and standard deviation = 2.5.
- where appropriate. 1. Experience has shown that the seeds from a certain variety of orchid have a 75% chance of germinating when planted under normal conditions. Suppose n seeds are planted, and let X be the random variable that counts the number of seeds that germinate. (a) What type of random variable is X? Indicate both the type and the appropriate parameters using the "~" notation. Write down the pmf of X, and do not forget to indicate the range of values that x can take on. X ~ px (x) = = (b) What is the minimum value of n so that the probability of at least five of the seeds germinating is at least 90%?A random sample of size n1 = 15 is selected from a normal population with a mean of 75 and a standard deviation of 9. A second random sample of size n2 = 9 is taken from another normal population with mean 69 and standard deviation 15. Let X1 and X2 be the %3D two sample means. Find: (a) The probability that X - X2 exceeds 3. (b) The probability that 4.9 < X – X2 < 5.9. Round your answers to two decimal places (e.g. 98.76). (a) i (b)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(15The random variable X takes value O with probability, and value 1 with the remaining probability. The random variable Y takes value 100 with probability and value 101 with the remaining probability. What is the relation between the variance of X and the variance of Y? Var[X] Var[Y] "Let x be a continuous random variable that is normally distributed with a mean of 123.5 and a standard deviation of 12.3. FInd the probability that x assumes a value between 97.9 and 113.2.A random sample of size n₁ = 14 is selected from a normal population with a mean of 76 and a standard deviation of 7. A second random sample of size n₂ = 9 is taken from another normal population with mean 71 and standard deviation 11. Let X₁ and X₂ be the two sample means. Find: (a) The probability that X₁ – X₂ exceeds 4. 1 2 (b) The probability that 4.3 ≤ X₁ – X2 ≤ 5.6. Round your answers to two decimal places (e.g. 98.76). (a) i (b) iA random variable Z is normally distributed with mean and variance are 0 and 1 respectively. Determine the probability of P( √2The profits of a mobile company are normally distributed with Mean of R.O (17x 10) and standarddeviation of R.O (17).a. Find the probability that a randomly selected mobile has a profit greater than R.O ((17x10) +10)Recommended textbooks for youA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSONA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON