A. Suppose the lengths of human pregnancies are normal distributed with u= 266 d

Question

A. Suppose the lengths of human pregnancies are normal distributed with u= 266 days and
o=16 days. Complete parts (a) and (b) below.

The figure to the right represents the normal curve with =266 days and o=16 days.
The area to the left of Upper X =245 is 0.0947. Provide two interpretations of this area.

1). Provide one interpretation of the area. Select the correct coice below and fill i the answer
boxes to complete your choice.

A. The proportion of human pregnancies that less than _______ days is _____.
B.The proporation of human pregnancies that last more than ___ days is ____.

2). Provide a second interpretation of the area. Select the correct choice below and fill in the answer boxes to complete your choice.

A. The probability that a randomly selected human pregnancy lasts less than _______ days is _____.
B. The probability that a randomly selected human pregnancy lasts more than _______ days is _____.

B. The figure to the right represents the normal curve with =266 days and o=1616 days.
The area between x=235 and x=300 is 0.9569.

1. Provide one interpretation of the area. Select the correct choice below and fill in the answer boxes to complete your choice.

A. The proportion of human pregnancies that last less than _______ days is _____.
B. The proportion of human pregnancies that last between _______ days is _____.

2. Provide a second interpretation of the area. Select the correct choice below and fill in the answer boxes to complete your choice.

A. The probability that a randomly selected human pregnancy lasts less than _______ days is _____.
B. The probability that a randomly selected human pregnancy lasts between _______ days is _____.

Explanation / Answer

PART I.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 266
standard Deviation ( sd )= 16
a.
P(X < 245) = (245-266)/16
= -21/16= -1.3125
= P ( Z <-1.3125) From Standard Normal Table
= 0.0947
the proportion of human pregnancies that less than 245 days is 9.47%
b. The probability that a randomly selected human pregnancy lasts more than 245 days is
= 1 - 0.0947
= 0.9053
=90.53%
PART II.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 235) = (235-266)/16
= -31/16 = -1.9375
= P ( Z <-1.9375) From Standard Normal Table
= 0.0263
P(X < 300) = (300-266)/16
= 34/16 = 2.125
= P ( Z <2.125) From Standard Normal Table
= 0.9832
P(235 < X < 300) = 0.9832-0.0263 = 0.9569
B. The probability that a randomly selected human pregnancy lasts between 235 and 300 days is 95.69%

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