PART VI. Consider the following discrete probability distribution. X P(x) 0 0.18 1 0.44 2 0.27 3 0.08 4 0.03 1. Calculate the mean or the expected value of X 2. Calculate the variance of the random variable, X 3. Calculate the standard deviation of the random variable, X
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- Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 1 2 3 4 5 Probability 0.259 0.293 0.244 0.150 0.037 0.017 ... (a) Find the mean, variance, and standard deviation of the probability distribution. The mean is (Round to one decimal place as needed.) The variance is (Round to one decimal place as needed.) The standard deviation is (Round to one decimal place as needed.) (b) Interpret the results. The mean is , so the average batch of 1000 machine parts has The standard deviation is so the typical number of defects in a batch of 1000 (Round to one decimal place as needed.)A random variable X has the following distribution: X P(X=x) -1 .3 0 1 .5 2 Let Y=X². Find the expected value of random variable Y. 0.38 0.6 5 4 O-1Answer the second two questions
- Find the variance and standard deviation of the probability distribution of a random variable X which can take only the values 3,5, and 7, given that P(3) 30 P(5) =, and P(7) = 20 13 %3D 2.Which of the following is a true statement? A. The expected value is the most frequently occurring value of the random variable. B. Variance is always >=0 C. Variance can be negative D. F(x)=P(X=x)Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 1 4 Probability 0.262 0.299 0.233 0.153 0.037 0.016 ... (a) Find the mean, variance, and standard deviation of the probability distribution. The mean is 1.5. (Round to one decimal place as needed.) The variance is (Round to one decimal place as needed.)
- Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 0 1 2 3 Probability 0.269 0.295 0.241 0.144 The variance is 1.5. (Round to one decimal place as needed.) (a) Find the mean, variance, and standard deviation of the probability distribution. The mean is 1.4. (Round to one decimal place as needed.) The standard deviation is 1.2 (Round to one decimal place as needed.) (b) Interpret the results. 4 0.038 The mean is so the average batch of 1000 machine parts has the typical number of defects in a batch of 1000 (Round to one decimal place as needed.) 5 0.013 0 The standard deviation is SOUse the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 1 2 4 Probability 0.264 0.289 0.238 0.156 0.037 0.016 ... (a) Find the mean, variance, and standard deviation of the probability distribution. The mean is 1.5. (Round to one decimal place as needed.) The variance is (Round to one decimal place as needed.)The table below shows the probability distribution of the random variable X. a. Find the mean of the random variable. b. Obtain the standard deviation o of the random variable. 1 P(X =x) 0.4 0.1 0.5 a. Find the mean of the random variable. (Round to two decimal places as needed.) b. Obtain the standard deviation o of the random variable. O = (Round to two decimal places as needed.) Enter your answer in each of the answer boxes. ? Le
- Find the mean, variance, and standard deviation of the following probability distribution by completing the tables below.A fitness center bought a new exercisemachine called the Mountain Climber. Theydecided to keep track of how many people used themachine over a 3-hour period. Find the mean, variance,and standard deviation for the probability distribution.Here X is the number of people who usedthe machine. X 0 1 2 3 4 P(X) 0.1 0.2 0.4 0.2 0.1