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Supposed four tiles numbered 1,2 ,3, and 4 are in a jar. A tile is picked and returned in the jar 15 times. The results are as follow: Number of times picked From the results, the average number per pick would be computed by: 1(2) + 2(4) + 3(8) + 4(1) x: 15 _2+8+24+4 15 = 2.53 This means that for every tile picked from thejar, the number in the tile is in average 2.53. This may not be a possible result of any individual yield or outcome, but this is very important measure in statistics. If we rewrite the calculation separating the tile number from the probability of each based on the based on the results, the computation would be: z=1(:.)+2(a+se)+4(:.)=2-ss 3 Note that the value 'tis'' and 115 are the probabilities of each tile numbered 1,2 ,3, and 4, respectively, based from the results. The value 2.53 in the example above is called expected value, mathematical expectation, or mean of the discrete random variable defined. Formula for the Mean of the Probability Distribution The mean of a random variable with a discrete probability distribution is: u = X1 0 P(X1) +X2 o P(X2) +X3 - P(X3) + "-,Xn I P(Xn) or u = Z X - For) Where X1,X2,X3 ,Xn are the values of the random variables X; and PUG); P(X2); P(X3); .P(Xn). ,P (Xn) are the corresponding probabilities. Grocery Items The probabilities that a customer will buy 12.3.4, or 5 items in a grocery store are 3112 a . . . E ,E , ' E' and 5' respectively. What IS the average number of Items that a customer will buy? Solution: 1. Construct the probability distribution for the random variable X representing the mlmhar nf item that Solution: Steps Solution 1. Construct the probability Number of items Probability distribution for the random X P(x) variable X representing the 1 3 number of item that the 10 customer will buy. 2 1 10 3 To 4 2 10 5 3 10 2. Multiply the value of the random Number of Probability X.P(X) variable X by the corresponding items P(x) probability. X 1 2 3 6/ W6 NG-6 -5/W 4 5 3. Add the results obtained in steps Number of Probability X.P(X) 2. items P(x) X 1 w 10 10 2 IN TO 3 TO 10 4 2 10 10 5 w 15 10 [ x . P (X ) = =3.1 So, the mean of the probability distribution is 3.1. This implies that the average number of items that the customer will buy is 3.1. Example: The probabilities that a surgeon operates on 3,4,5,6,or 7 patients in any days are 0.15, 0.10, 0.20, 0.25, and 0.30, respectively. Find the average number of the patients thatExample: The probabilities that a surgeon operates on 3,4,5,6,or 7 patients in any days are 0.15, 0.10, 0.20, 0.25, and 0.30, respectively. Find the average number of the patients that a surgeon operates on a day. Solution: Steps Solution 1. Construct the probability distribution for the Number of Patients Probability random variable X X P(X) representing the number 3 0,15 of patients that a surgeon 4 [110 operates on a day. 5 020 6 0.25 7 0.30 2. Multiply the value of the random variable X by Number of Probability X -P(X) corresponding probability. Patients PlX) X 3 0.15 0.45 4 0.10 0.40 5 0.20 1.00 6 0.25 1.50 7 0.30 2.10 3. Add the results obtained in Step 2. Number of Probability X IPlX) Patients PlX) X 3 0.15 0.45 4 0.10 0.40 5 0.20 1.00 6 0.25 1.50 7 0.30 2.10 2x - P(X) = 5.45 So, the average number of patients that a surgeon will operate in a day is 5.45. Explore your understanding: Given the values of the variables X and Y, evaluate the following summations. X1=5 X2=3 X3=4 X4: 0 Y1=1 Y2=0 Y3=2 Y4=2 Find: 1 2X 2.21? 3.ZX+Y 4.22XY Firm - Up your knowledge: h/ Flrm - Up your knowledge: Complete the table below and find the mean of the following probability distribution. 1. x Pix) 1 1 5 1 6 7 3 11 5 1 15 7 21 l + Deepen your understanding: Solve the problems. The following table gives the probabilities that a probation officer will receive 0,1, 2,3,4,or 5 reports of probation violations on any given day. Number of Violations Probability Pix) Transfer: Find the mean of the probability of the random variable X, which can take only the values 2, 4,5, and 9, given that P(2) = 290 PM) = % 9(5) = g and me) = 130. x-Po