We assume that the temperature during the month of March follows a normal distribution with an expectation of 20 and a variation of 11.0889, noting that (P < 2)=9772 and P( Z< 0.33)=0.6293 and P (Z< 1) = 0.8413, then P ( 21.11

A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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We assume that the temperature during the month of March follows a
normal distribution with an expectation of 20 and a variation of
11.0889, noting that (P< 2)= 9772 and P(Z< 0.33)=0.6293
and P (Z< 1) = 0.8413, then P ( 21.11 <y < 26.66)is??
0.7439 (7
0.3479 (
0.4397 (i
Transcribed Image Text:We assume that the temperature during the month of March follows a normal distribution with an expectation of 20 and a variation of 11.0889, noting that (P< 2)= 9772 and P(Z< 0.33)=0.6293 and P (Z< 1) = 0.8413, then P ( 21.11 <y < 26.66)is?? 0.7439 (7 0.3479 ( 0.4397 (i
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