Question

The answers are in red and thanks in advance

The average weight of a professional football player in 2009 was 255.6 pounds. Assume the population standard deviation is 30 pounds. A random sample of 31 professional football players was selected. Complete parts a through e. a. Calculate the standard error of the mean G = 5.39 (Round to two decimal places as needed.) b. What is the probability that the sample mean will be less than 242 pounds? PX<242) = 0.0058 (Round to four decimal places as needed.) c. What is the probability that the sample mean will be more than 240 pounds? P(x>240) = 0.9981 (Round to four decimal places as needed.) d. What is the probability that the sample mean will be between 262 pounds and 269 pounds? P(262 sxs 269) = 0.1110 (Round to four decimal places as needed.) e. Identify the symmetrical interval that includes 95% of the sample means if the true population mean is 255.6 pounds. 245 sxs 266 (Round to the nearest pound as needed.)

Answer 1

Some answers are a bit different in decimal because you might have used 5.39 for all other problems, but I’ve taken 30/31^0.5 everywhere.

POX-29- La 3013) Q. @ 52=0/16 = 307531 - 15.39) Opaž<242) Plze 242-255.6) - PIZZ-2.52) 210.0059) © Pl#>240)- P(z>246-255.6 ) - P(Z >-2,90)-10.9981 30/31 ) @P(26250 € 269)= P( 262-255.b çzs 269.255) - P(1.19425 2.49) - 0.9936-08830-10.906 © P(-20% Z <7)= 0.95 :) Zo=1.96; +2556 €[-1.96,196] 255.6-1.96*3 *32556+1.96* 30/331 ..[2454 X4 2.2 30 Nai