Question

Use the following information for all remaining problems on this worksheet. A physical fitness association is including the mile run in its secondary school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 440 seconds and a standard deviation of 40 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.

probability is 0.0107

Answer 1

Let x denotes time taken by student for a given event XN N(M = 440 8² 402) Probabiling that a randomly selected boy in secondary student can run the mile in less than 34.8. see. P (82348) = P(x2440 3487449). 40 .:= p C. zc -2.3 ) L = 0.0107 -2.3 This probabiling i ů for a randomly selected shedent If sample of 100 is selected then expected number of boys who run in less than 348 see i given by Expected no. of skeefents = 100 x 0:0107 . boys .: 1.04 Probability that sandomly selected boy in secondary school will take longer than 345 see. to non me mile s p (x>345) -2.315 = p (2-2.375.) 1-PC22 -2.375) = 1-0.008.7 5.0.9913

Graph represent the area for the probability el mil. take a randomly selected boy longer than 345 seands to run a mile ej. . 1 l = 440 345 of Between to nen what times do we expect the mile. 95% ; the boys. 0.95 P (a < x < 2) = 0.95 .. . -2, x2 P (memoto che foto e moto ):0.95 po 240 cze M 1,6040 ) - 0.95 90 A correct option is between 30116 g 5.18.4 see as follow. PC 361.60440 cz < 546.4)= { (-196<2<1.96). - 2P(2<1.96) - 2X0.4750 . E oigs

4472 Probability that boy nun a mile between 372 seconds = p (372 < X 472) EP (-1.7.22 0.8) : a. p (-1.72220) +P (0 < 220.8). -1.7 0.8 Ž0.4554 +0.2881 = 0.7435 percentage of the boys would be expected to run the mik in between 312 & 472 seconds = 100 x P (872 CX 1974 = 100 x 0.79.35 = 14.35 ... correct option is 94.4% . correct graph is shaded region & penen fage gives the probability | M = 440 472 312