μ=261 days
σ=20 days
Z= (Xbar-mu)/(sd) is the formula for normal distribution
a)Z=(296-261)/20 =1.75
P(Z>1.75) =0.04 (From Z curve find area under the curve for z value more than 1.75)
Hence we can say that 4% is the proportion of pregnancies lasts more than 296 days
b) Z=(256-261)/20 = -0.25
& Z=(266-261)/20 = 0.25
P(Z>-0.25 & Z<0.25) =19.74%
c) The probability that a randomly selected pregnancy lasts no more than 251 days
Z=(251-261)/20 = -0.5
P(Z>-0.5)=69.15%
d) = (231-261)/20 = -1.5
P(Z<-1.5)=6.68%
Hence we can say that the probability of premature babies are 6% which is not very unusual.
Hope the above answer has helped you in understanding the problem. Please upvote the ans if it has really helped you. Good Luck!!
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