Question

0.09/1 points Previous Answers My Notes Consider babies born in the "normal" range of 37-43 weeks gestational age. A paper suggests that a normal distribution with mean = 3500 grams and standard deviation d is a reasonable model for the probability distribution of the continuous numerical variable x-birth weight of a randomly selected full-term baby r ams (a) What is the probability that the birth weight of a randomly selected full-term baby exceeds 4000 97 (Round your answer to four decimal places.) (b) What is the probability that the birth weight of a randomly selected full-term baby is between 3000 and 4000 g? (Round your answer to four decimal places.) (c) What is the probability that the birth weight of a randomly selected term baby is either less than 2000 g or greater than 5000 g? (Round your answer to four decimal places.) (d) What is the probability that the birth weight of a randomly selected full-term baby exceeds 7 pounds? (Hint: 115 453.59. Round your answer to four decimal places) (e) How would you characterize the most extreme 0.1% of all full-term baby birth weights (Round your answers to the nearest whole number) grams and those less than grams The most extreme 0.1% of birth weights consist of those greater than (1) Ifx is a random variable with a normal distribution and as a numerical constant (a 0), then y axalso has a normal distribution. Use this formub to determine the distribution of full term baby birth weight expressed in pounds(shape, mean, and standard deviation), and then recalculate the probability from part() (Round your answer to four decimal places.)

Answer 1

* Given b } mean (M = 3500 grams S.D(6)=588 gram Let x denotes the birth weight of a randomly selected full term baby. PCX Junoot Cat - 4000-3500 ) or 5188. M ap (z > 0.85) = 1- P(Z <0.85) ii- 0.8023 b (b) al Yooo 3500 588 PC 3000 c X24000) = p (3000-3500 exclosos) = P(-0.85.2.240.85) =P(z<-0.85) - P(z>0.85) = 2(0.8023) – 1 =0.6046.

(C) P(x (2000) + P(x > 5000) = PC** < ) + PCX H Seco - 3500) 588 = P(Z <-2.55) + P(z>2:55) = 0.0054 + 1 p (252.55) - 0.0054 +0.0054 =T0.0108 (d 1 pound = 453.592 grams ...7 pounds & 3175 144 g i. 3175.144-35500 588 P(x> 3175-144)=P(et 1 = P(Z >-0.55) = 1- P(26-0.55) = 1-0.2912 = 0.7088