Question

Assume that the duration of human pregnancies can be descibed by a Normal model with mean 267 days and standard deviation 13 days. a) What percentage of pregnancies should last between 270 and 281 days? b) At least how many days should the longest 30% of all pregnancies last? c) Suppose a certain obstetrician is currently providing prenatal care to 46 pregnant women, Lety represent the mean length of their pregnancies. According to the Central Limit Theorem, what's the distribution of this sample mean, y? Specify the model, mean, and standard deviation. d) What's the probability that the mean duration of these patients pregnancies will be less than 263 days? a) The percentage of pregnances that should last between 270 and 281 days is Round to two decimal places as needed.)

Answer 1

Let , $Y\to N(\mu=267,\sigma=13)$

a)

$P(270<Y<281)=P(\frac{270-267}{13}<\frac{Y-\mu}{\sigma}<\frac{281-267}{13})$

$=P(0.23<Z<1.08)$

$=P(Z>0.23)-P(Z>1.08)$

$=0.40905-0.14007$ ; From standard normal probability table

$=0.2690=26.90\%$

b)

Since , $P(Y>y)=0.30$

$P(\frac{Y-\mu}{\sigma}>\frac{y-267}{13})=0.30$

$P(Z>\frac{y-267}{13})=0.30$ ...........(1)

From standard normal probability table

$P(Z>0.52)=0.30$ .............(2)

From (1) and (2) , We get ,

$\frac{y-267}{13}=0.52$

$y=0.52*13+267=273.76$

(C) Since , n=46

If $Y\to N(\mu,\sigma)$ , then $\bar{Y}\to N(\mu_{\bar{Y}},\sigma_{\bar{Y}})$

Therefore , the sampling distribution of sample mean is normal distribution

Where , $\mu_{\bar{Y}}=\mu=267$

$\sigma_{\bar{Y}}=\sigma/\sqrt{n}=13/\sqrt{46}=1.9167$

d)

$P(\bar{Y}<263)=P(\frac{\bar{Y}-\mu_{\bar{Y}}}{\sigma_{\bar{Y}}}<\frac{263-267}{1.9167})$

$=P(Z<-2.09)$

$=P(Z>2.09)$

$=0.0183$ ; From standard normal probability table