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- 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%?CALL CENTERExcopt for PANAMA) 1-800-SAMSUNG (1-800-726-75 6. For a population with a mean of u = 100 and a standard deviation of o = 20, a. Find the z-score for each of the following X values. X = 108 X = 115 X = 130 %3D %3D %3D X = 90 b. Find the score (X value) that corresponds to each of the following z-scores. X = 88 X = 95 z = -0.40 z = -0.50 z = 1.80 z = 0.75 Z = 1.50 Z = -1.25can you solve this for me by showing step by step for each question? Acme Food manufactures and sells bags of doughnuts whose weights follow a normal distribution with a mean of 400g and standard deviation 8g. a) Find the probability that a bag of doughnuts weights less than 410g b) Find P (X>412) c) Find P (395 < X < 404) d) The company can be prosecuted if the bags are less than 380g, if the company sells 50 bags per day find the expected number of bags under this weight. e) On a typical day 1.22% of bags are rejected as they are too heavy. Find the rejection weight.
- T1.Q3) A university class ckntained 60 female students. they were asked to report their hieghts (in cm), with the following results. n=60 mean=163.39 standard deviation=8.68 1) within what range would you expect 95% of the heights to lie? 2) Find the z-score of Jenny, a female student whose height is 155.22cm. Approximately how many students are shorter than Jenny? 3) Tina has a height of 195cm. what is her z- score? Approimately how many students in the class do you think would be taller than Tina?6.- Assume that cans of Coke are filled so that the actual amounts have a mean of 12.00 oz and a standard deviation of 0.11 oz. a). Find the probability that a sample of 36 cans will have a mean amount of at least 12.19 oz. P( > 12.19) = b) Find x0 such that P(11.95 < X < x0 ) = 0.99Let a population consist of the values 7 cigarettes, 21 cigarettes, and 22 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Sample Mean Absolute Deviation {7,7} __ {7,21} __ {7,22} __ {21,7} __ {21,21} __ {21,22} __ {22,7} __ {22,21} __ {22 ,22}
- Suppose that the weights of 600 male students are normally distributed with mean u = 60 kg and a = 6 kg then the number of students with weights between 54 and 66 is 190 None 410 505Let x be a random variable that represents the weight of a lion in a certain region of Africa. Assume that x is normally distributed with a mean = 260 lbs . With standard deviation =30 lb Please include the graphs for questions A and B1)When can a Standard Deviation for set of 5 numbers be zero? In other words, what kind of relationship must exist among all numbers in order for Standard Deviation to be zero Hint: Standard Deviation E(X¡ - Mean)²/(n – 1)
- A random sample of size n1 =15 is selected from a normal population with a mean of 76 and standard deviation of 9. A second random sample size n2 =10 is taken from another normal population with mean 71 and a standard deviation 13. Let x barı and x bar2 be the sample means. Find: a. The probability that xbar1 – xbar2 exceeds 4.5 b. The probability that 3.5 s xbar1– xbar2 s 5.5Let a population consist of the values 11 cigarettes, 12 cigarettes, and 22 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Mean Absolute Deviation Sample {11,11} {11,12} {11,22} {12,11} {12,12} {12,22} {22,11} {22,12} {22,22} (Type integers or decimals rounded to one decimal place as needed.) Budget Pi nd Pow....P Enter your answer in the edit fields and then click Check Answer. Clear All Check Answer partsA student takes a true-false test that has 10 questions and guesses randomly at each answer. LetX be the number of questions answered correctly. c)The mean and stand deviation of X (show work and round final answer to 2 decimal places) Mean: Standard deviation =
