Question

incubation for chicks has a normally distributed mean of 24 days and standard deviation of 3 days, if 1000 eggs and being incubated how many chicks do we expect to hatch between 18-30 days, 21-27 days, 21 days or fewer and 15-33 days

Answer 1

1) P(18< X 30) Normal Distribution, 24, ơ-3, we convert this to standard normal using 18 (24) _-2.00 Zl= 30-(24) -2.00 P(-2.00< Z< 2.00) Area in between -2.00 and 2.00 0.97720.02280.9544 1 9:9772 2.00 -2.00 P(18 < X < 30) = P(-2.00 < Z < 2.00) = P(Z < 2.00)-P(Z <-2.00) = 0.9772-0.0228-0.9544 P(Z < 2): in a z-table having area to the left of z, locate 2 in the left most column. Move across the row to the right under column 0.00 and get value 0.9772 P(Z<-2): in a z-table having area to the left of z, locate -2 in the left most column. Move across the row to the right under column 0.00 and get value 0.0228 The expected number is xp 1000 0.9544-954 (or 954.499736103642

using technology 2) P(21 < X 27) Normal Distribution, 1-24, ơ-3, we convert this to standard normal using 21- (24) 一一1.00 27 (241.00 P(-1.00Z1.00) Area in between -1.00 and 1.00 0.8413-0.15870.6826 0.8413 0.1587 1.00 P(21 < X < 27) = P(-1.00 < Z < 1.00) = P(Z < 1.00)-P(Z <-1.00) = 0.8413-0.1587= 0.6826 P(Z <1): in a z-table having area to the left of z, locate 1 in the left most column. Move across the row to the right under column 0.00 and get value 0.8413 P(Z<-1): in a z-table having area to the left of z, locate -1 in the left most column. Move

The expected number is- nxp 1000*0.6826683 (or 682.689492137086 using technology) 3) P(X< 21) Normal Distribution, μ-24, σ-3, P(X < 21)-Area to the left of 21 we convert this to standard normal using z - 21 _ (24)-- 00 0.1587 P(X < 21) = P(Z <-1.00) = 0.1587 (from z-table) P(Z<-1): in a z-table having area to the left of z, locate -1 in the left most column. Move across the row to the right under column 0.00 and get value 0.158 The expected number is = n × p = 1000 *0.1587 = 158.655253931457000 using technology) 159 ~ 159 (or

4) P(15< X 33) Normal Distribution, 24, ơ-3, we convert this to standard normal using 15 (24) -3.00 33- (24) 33.00 P(-3.00Z< 3.00) Area in between -3.00 and 3.00 0.9987-0.0013 0.9974 0.0013 0.99 3.00 -3.00 P(15 X<33) P(-3.00< Z< 3.00) = P(Z < 3.00)-P(Z <-3.00) = 0.9987-0.0013-0.9974 P(Z < 3): in a z-table having area to the left of z, locate 3 in the left most column. Move across the row to the right under column 0.00 and get value 0.9987 P(Z-3): in a z-table having area to the left of z, locate -3 in the left most column. Move across the row to the right under column 0.00 and get value 0.0013 The expected number is nxp 1000 0.9974-997 (or 997.3002039367400 using technology)