Suppose the durations of human pregnancies follow a mound-shaped distribution wi

Question

Suppose the durations of human pregnancies follow a mound-shaped distribution with a mean of 266 days and a standard deviation of 16 days. a. Approximately what percentage of human pregnancies last between 250 and 282 days? b. Between what two values will the durations of 95% of all human pregnancies fall? Convert these values to months (with 1 decimal place), assuming the typical month has 30 days. From to months. Suppose also that the durations of horse pregnancies follow a mound-shaped distribution with a mean of 336 days and a standard deviation of 3 days. c. Is a horse or a human more likely to have a pregnancy that lasts within ±6 days of its mean? Explain briefly. A . In fact 95% of all pregnancies will last within roughly ± 6 days of the mean because the standard deviation for pregnancies is days.

Explanation / Answer

Ans:

mean=266

std dev=16

a)

z(250)=(250-266)/16=-1

z(282)=(282-266)=1

P(-1<=z<=1)=P(z<=1)-P(z<=-1)=0.8413-0.1587=0.6826

b)

z value=+/-2

lower limit=266-2*16=234 days=7.8 months

upper limit=266+2*16=298 days=9.9 months

c)z value for human=+/-6/16=+/-0.375

z value for horse=+/-6/3=+/-2

P(-0.375<=z<=0.375)=0.6462-0.3538=0.2923

P(-3<=z<=3)=P(z<=3)-P(z<=-3)=0.99865-0.00135=0.9973

Hence,horse is more likely to have a pregnancy that lasts within ±6 days of its mean.

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