Question

The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 308 days or longer. b. If the length of pregnancy is in the lowest 4%, then the baby is premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. a. The probability that a pregnancy will last 308 days or longer is (Round to four decimal places as needed.)

Answer 1

Let X be the length of pregnancy in days

X N (269, 152

then

X-269 15N 0.11.

a) To find

$P(X\geq 308)$

308 - 269 = P(:>

= P( z> 2.6)

= 0.0047  (from z table)

Probability that pregnancy will last 308 days or more 0.0047

b) To find C such that

P( X <C ) = 0.04

$\Rightarrow P(z<\frac{C-269}{15})= 0.04$

From z table

P( z< -1.76 ) =0.04

Thus

$\frac{C-269}{15}=-1.76$

$\Rightarrow C=242.6$

Coverting to whole number , C =243

Thus length that separates premature babies from those who are not = 243 days

2. Let X be the overhead reach of adult females in cm

$X\sim N(202.5,8.6^2)$

then

$z=\frac{X-202.5}{8.6}\sim N(0,1)$

(a) To find

$P(X\geq 215.90)$

$=P(z>\frac{215.90-202.5}{8.6})$

= P( z> 1.56 )

= 0.0594   (from z table)

Probability that reach distance is greater than 215.90 cm is 0.0594

b)

Let $1586409052542_blob.png$ be the average of 15 randomly selected distances

X Normal
with mean = 202.50 cm and standard error = $8.6/\sqrt{15}= 2.22$

Then

$z=\frac{\bar{X}-202.5}{2.22}\sim N(0,1)$

$P(\bar{X}>201.20)$

$=P(z>\frac{201.2-202.5}{2.22})$

= P( z > -0.59)

=0.7224 (from z table)

Probability = 0.7224

c) The sampling distribution of sample mean $\bar{X}$ follow Normal distribution in following two conditions

(i) If sample size is large even though parent population is not normal

(ii) If the parent population is normal even though sample size is small

Here parent population of distance is normal , thus even though sample size is small the sample mean follow Normal .