Question

Please help with explanations

(a) In a Lab Experiment involving inorganic syntheses of molecular precursors to organometallic ceramics, the final step of a five-step reaction involves the formation of a metal- metal bond. The probability of such a bond forming is p = 0.30. Let X equal the number of successful reactions out of n = 25 such experiments. Find the following probabilities: O (ii) (ii) (iv) (v) (vi) P (X is at most 5) P (X is at least 6) P (X = 6) Find the mean of X = u = np Find the variance of X =g2 np. q = 1-P Find the standard deviation of x = 0 3 points 3 points 3 points 2 points 3 points 2 points (b) Joan has noticed that the probability distribution for x = number of students in line to use the campus ATM machine when she shows up to use it is shown below: X 0 1 2 3 4 p(x) 10 .10 40 .30 10 (0) What is the probability that there will be no student in line when Joan shows up? 3 points What is the probability that there will be no more than one student in line when Joan shows up? 3 points (i) What is the mean = Expected value of x = E(x)= 3 points

Answer 1

su 2. succe (a) xi equal numbet of successful reactions Whete, forming a bond is success and mot forming is failure. So, probability of forming a bond = p=0.30 probability of mot forming a bond -92 tomto Numbee of preperiments = =25 Here, xn binomial ( =25, p=0.30) otting (0) P(x is at most 5) = Px35) = P(X=0) +...+P[X=5) = 25C66030)*0.75* = 10. 1935 (ii) P( X 15 at least 6) = P(X716) = 1- P(X<5) = I- 0. 1935 =10.8065 - Y xn-2 formula: P(x=x) = cm P 9 Izo

472 (6) PIX=6) = 25C(0.56C07266 =10.1442 (fv). Mean=le=mp = 25%0.3 = 7.5 le average of successful bond Roomation expediments is 8.5 (1) valliance (x) = 8?= npq = 25X03 X0.7 -15.25 (vi) Standeed deviation, sd(x) = J vaziance (X) = 15.25 =12.2913

B Giyen, Xe Number of student in bīne 1. When Joan shows up. () P (there will be no student] = p [xzo] =) 0.40] <(Using given table) (ii) p [no more than 4] = p[x51] =P(x =o) +P(x=1). = 0.50 +0.40 € (sing table given) =10.20 (1) Mean (X) = Expected Value ofix = E (x) = 3 X PCx9 = [0x0.1] + [1 x0.-1]+[2x0.4] + [3X 0.3] + [4x0.j] = 0 +0.170.8 +0.9 +0.4 = [2.2) le. Average numbee of shedent in line is 2.27