Question

4. 19. Bricks' weights are independently distributed as a normal distribution with mean of 110 and standard deviation of 2.5. If a total of n= 20 bricks is calculated, what is the probability that the total weight of those 20 randomly selected bricks is greater than 2220? (on calculator take upper to be larger than 1000, say 100000) Draw curve, 4 places past decimal:

Answer 1

Bricks kleights are independently distributed as normal Given population mean Cul - 110 Given Standard deviation (o) = 2.5 n total kleight of those ao randomly selected } = ? Pl bricks is greater than aaao. Sample mean for those 20 brucks (I) = 2220 - 11 A P(X>111) = ? . Now, in this Suitation we use Standard normal Karate z P ( 4 > -4) - P(Z >" ) - P(Z >0.4) a Plz>0.4) - P(230.4) P(Z <0.4) PC250-4) too 0 0.4. : P(Z > 0.4) = 1-P(2204) From standard normal tables, Pl 504) = 0.6554 AP(2 704) = 1-0.6554 = 0.3446 c. probability that the total kleight of those 20 randomly Selected bricks is greater than 2920 = 0.3446