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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 鈥減regnancy lasts between 325 and 345 days鈥� in terms of X. Then find

this probability.

b. Find the value of c such thatP(X鈮�c)=0.20

Short Answer

Expert verified

a) The probability for the event 鈥減regnancy lasts between 325 and 345 days鈥� in terms of X is89.96%.

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

Step by step solution

01

Step 1. Given information. 

We have given values of mean 渭=336standard deviation 蟽=6andx=325or345

Let Xrepresent the length of a randomly selected horse pregnancy.

02

a) Step 2. To find the probability. 

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

z=x-渭蟽=325-3366鈮�-1.83z=x-渭蟽=345-3366=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%

03

b) Step 3.To find the value of c.

We have, P(X鈮�c)=0.20

P(notA)=1-P(A)P(x<c)=1-P(x鈮�c)P(x<c)=1-0.20P(x<c)=0.80

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

z=x-渭蟽=x-3366c-3366=0.84c-336=0.84脳6c=336+0.84(6)c=341.04

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