Question

-/9 points BBUNDERSTAT126.5.011. Suppose x has a distribution with 10 and a = 5. (a) If a random sample of size n = 41 is drawn, find and RX10 X 12). (Round , to two decimal places and the probability to four decimal places.) 07 P(10 SX S12) - and P(10 SX512). (Round to two decimal places and the probability to four decimal places.) (b) If a random sample of size n = 73 is drawn, find x = 0 = P(10 SXS 12) = (c) Why should you expect the probability of part (b) to be higher than that of part (a? (Hint: Consider the standard deviations in parts (a) and (b).) The standard deviation of part (b) is --Select- part (a) because of the Sele sample size. Therefore, the distribution about is Select

Answer 1

given U = 10 ay Sample size (n) = 41 inux We know that ex=0 Mg = 10) We know that of olo =) Srul = 0.7808 08-0.7808) " y losť 212) 5 olra < x-1 < 12-10 > pl 10-10 10.7801 0.7505 =) pcos72.5614) =) pcz<2.5614) - przco) -> 0.9798-018 IP Clocł sig=0.u798)

by n=73 (1) My excel we know that -lle 10 (1) we know that of an => 5 73 1) 73773 = 5773 73 = 0.5852 loa=0.5852 p C10 < x <12) » POR CREATE c10-10 = p losoa eta 0.5852 »plosza 2%.,852) 1) Plos?? 3.1476 ) > P(Z <3-1476- P(Z <o) =) 0.99aa-or Z < 12-10 sp ( 852

=) 0.4992 lloc XL 12) = 0.4992 c7 from the alb The probability of bis higher than probability b. . . The standard deviation of pouts is [0.5854 because of scumple size 73.