X 1 2 expecta ctation fo P(x) 0.23 0.41 0.19 0.17 3 4 OA) 1.0 O this probability distribution?
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- The random variable X has the following probability distribution. -1 0 P(X=x) 0.20 0.50 0.30 Calculate E[X]. Round to two decimal places.Rion 15 Given a probability distribution tanswered d out of 2.00 question 0 1 2 3 4. P(x) 2k 0.15 0.2 0.05 Find the value of k. • 1 0.3 • 0.6 0.2I. Which of the following are discrete probability distributions? (Yes or No) 1. 4 P(x) 0.10 0.20 0.25 0.40 0.05 2. 2. 3 4 5 P(x) 0.05 0.25 0.33 0.28 0,08 3. 3 4 5 P(x) 0.08 0.25 0.34 0.31 0.04 4. 2 3 4 5 P(x) 0.03 0.22 1.01 0.23 0.02 5. 1 2 3 4 P(x) 0,05 0.27 0.34 0,28 0.06 6. 3 4 5 P(x) 3 3 10 10 10 7. 3 4 P(x) 15 15 8. 3 4 5 P(x) 25 25 25 25 9. 3 4 P(x) 7. 1 10 10 20 20 10. 2. 3 4 5 P(x) 0.212 0.113 0,125 0.224 0.306
- 4:48 Mid-Statistics Exam First Semester (Sta.. Section 5. Arrivals of customers at a check counter follow a uniform distribution. It is known that, during a given 30- minute period, one customer arrived at the counter. Find the probability that the customer arrived during the last 5 minutes of the 30- minute period? * (2/2 Points) 0.0333 0.2000 0.0500 0.0300 0.1666 v None of the Above Activity Chat Teams Assignments More More EditPlease answer this statistic question0.15 0.4 0.25 A B (i) Calculate the conditional probability that A occurs, given that B has occured. (ii) Calculate P(A'/B)
- X P(x) 0 0.15 1 1 0.25 2 0.2 3 0.4 Find the standard deviation of this probability distribution. Give your answer to at least 2 decimal places Submit Question acer1 point Let X denotes the number of patients who die at a certain hospital in a day. The following table gives the probability function of X. 3 4 P(X = x) 0.3 0.25 0.25 0.15 0.05 The expected number of deaths per day in the hospital is: O 1 2.4 1.7 3.3 poirAn unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, XX, is as follows: xixi 1 2 3 4 5 6 Total P(X=xi)P(X=xi) 0.15 0.16 0.33 0.2 0.11 0.05 1 The expected value is computed, E[X]=3.11E[X]=3.11. To find the variance and standard deviation the following table was set up: xixi P(X=xi)P(X=xi) (xi−E[X])2(xi-E[X])2 (xi−E[X])2P(X=xi)(xi-E[X])2P(X=xi) 1 0.15 (1−3.11)2=(−2.11)2=4.4521(1-3.11)2=(-2.11)2=4.4521 4.4521⋅0.15=0.66784.4521⋅0.15=0.6678 2 (2−3.11)2=(−1.11)2=1.2321(2-3.11)2=(-1.11)2=1.2321 1.2321⋅0.16=0.19711.2321⋅0.16=0.1971 3 0.33 (3−3.11)2=(−0.11)2=(3-3.11)2=(-0.11)2= 0.0121⋅0.33=0.0040.0121⋅0.33=0.004 4 0.2 (4−3.11)2=(0.89)2=0.7921(4-3.11)2=(0.89)2=0.7921 0.7921⋅0.2=0.7921⋅0.2= 5 (5−3.11)2=(1.89)2=3.5721(5-3.11)2=(1.89)2=3.5721 3.5721⋅0.11=0.39293.5721⋅0.11=0.3929 6 0.05 8.3521⋅0.05=0.41768.3521⋅0.05=0.4176 E[X]=3.11E[X]=3.11 Total:…
