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JO Y:EY >> > For Q. 19 -23: The scores (grads) of students in computer lab are SI, S2, S1, S3, S2, $2, S4, S4,

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JO Y:EY >> > For Q. 19 -23: The scores (grads) of students in computer lab are SI, S2, S1, S3, S2, $2, S4, S4, S4, SI, S2, $3, $3, Ss, $2, $3, $1, SI, SI, $2, $3, S3, $5, Ss. Ss, $2, Ss, Ss and $2, where Si high score while Ss low score. Pa 19) The frequency distribution table of scores is A) Scores S1 S2 S3 SA Ss B) Scores S1 S2 S3 SA Ss Frequency 6 8 7 Frequency 6 8 7 4 5 C Scores S1 $2 S3 SA Ss D) Scores S1 S2 S3 SA Ss Frequency 6 8 7 4 6 Frequency 6 8 7 5 5 (20) The appropriate Pie chart that represent scores is A B $5; 4; 14% $1; 6: 21% $5;5; 17% 51; 6; 20% 54: 4; 14% 54; 4; 13% $2; 8; 27% $2; 8; 27% $3; 7; 24% $3; 7; 23%% C D $5; 6; $1: 6 $5: 5 $1: 6; 19% 19% 17% 21% 54: 3: $4: 4 10%% 13% $2; 8 26% $2: 8 53: 7: 28% 23% 24%% (21) The median of scores is A S B S2 C SS D Not exist (22) The mean of scores is AS B S C Ss D S (23) The mode of scores is A S B S C Ss D SA O>>> For Q. 58 - 60: For the following system, it can be assumed that components fail independently of one another. The probabilities of failure of the various components in the next year are indicated in the diagram above. 0.05 0.03 0.1 (58) The probability that the subsystem 0.03 fails is: (A) 0.0603 (B) 0.0123 (C) 0.0469 (A) 0.0785 (59) The probability that the subsystem 0.1 fails is: 0.1 Page (B) 0.01 (A) 0.02 (B) 0.2 (C) O.I (60) The probability that the whole system does not fail in the next year is: (C) 0.946111 (D) 0.469451 (E) 0.999715 (F) 0.468231 >> For Q. 61-63: Consider the system of figure below. Suppose that components are functioning. Independently and the probabilities that components C1 , C2, C3 , and C4 are functioning : 0.9 , 0.85 , 0.75 , and 0.7 respectively. C1 C3 (61) The probability that the system is functioning is (A) 0.765 (B) 0.888 (C) 0.725 (D) 0.35 (E) 0.466 (62) The probability that system is functioning given that component CI is not functioning is (A) 0.765 (B) 0.888 (C) 0.725 (D) 0.35 (E) 0.466 (63) The probability that system is functioning given that component C4 is not functioning is (A) 0.765 0.888 (C) 0.725 0.35 (E) 0.466 O

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