Chapter 6: Q 29. (page 373)

Horse pregnanciesBigger animals tend to carry their young longer before birth. The
length of horse pregnancies from conception to birth varies according to a roughly Normal
distribution with mean $336$ days and standard deviation 6 days. Let X = the length of a
randomly selected horse pregnancy.
a. Write the event “pregnancy lasts between 325 and 345 days” in terms of X. Then find
this probability.
b. Find the value of c such that $P (X \geq c) = 0.20$

Expert verified

a) The probability for the event “pregnancy lasts between 325 and 345 days” in terms of X is $89.96 %$ .

b) The units of the data value are days and thus $c = 341.04$

Step by step solution

We have given values of mean $μ = 336$ standard deviation $σ = 6$ and $x = 325 o r 345$

Let $X$ represent the length of a randomly selected horse pregnancy.

The z- score is the value decreased by the mean, divided by the standard deviation .

$z = \frac{x - μ}{σ} = \frac{325 - 336}{6} \approx - 1.83 z = \frac{x - μ}{σ} = \frac{345 - 336}{6} = 1.50$

Now,

$P (325 < X < 345) = P (- 1.83 < X < 1.50) = P (Z < 1.50) - P (Z < - 1.83) = 0.9332 - 0.0336 = 0.8996 = 89.96 %$

We have, $P (X \geq c) = 0.20$

$P (n o t A) = 1 - P (A) P (x < c) = 1 - P (x \geq c) P (x < c) = 1 - 0.20 P (x < c) = 0.80$

Thus , the corresponding z- score is $0.8 + . 04 = 0.84$ .

$z = \frac{x - μ}{σ} = \frac{x - 336}{6} \frac{c - 336}{6} = 0.84 c - 336 = 0.84 \times 6 c = 336 + 0.84 (6) c = 341.04$

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