Question

Babies born in singleton births in the U.S. have weights that are approximately normally distributed with mean 3.5 kg and standard deviation 0.8 kg. Regarding this population: (A) What is the weight of a baby in the 5th percentile? That is, find the weight such that 5% of babies weigh less. (B) Suppose babies lighter than the 5th percentile require special treatment for low weight. Suppose, further, that a hospital typically has ten newborn babies per day that may be regarded as randomly and independently chosen from the population. What is the probability that two of a day's ten babies are lighter than the 5th percentile? (A) mean of X (B) mean of Y (C) var(x) (D) var (y)

Please answer these questions with the use of a calculator and not Z-tables. Please provide all information necessary to arrive at the solutions.

Answer 1

X. o saby weight ~ N(4=3-5, 8=0.8) Ą Weight of 5 * fexcentile let the go Bo Here X-u No1). say Z = X-4 We want Ks. t P& <K),ous 2 98k) : 0.05 - P(XM 2K4) 2005 » P(Z < K-M) 2005 - ? [Z 76(K-4)] = 0.05. [z<uk 472 0.95 and 95th berceilile gestandend Normelis 1645 > 6 (K:4) = 1:645 (K = 8:184]

B p / out of to. are lighter) let R: # of babies lighter than sta percentile autolla ha Binomial (nb) - nclo, p = 0.05 » P (8-2) = loc p²0-6) - =[0.0746 Required probability that two of lochildove leghitel than the percentilo '