Question

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$P(X\ge c)=0.20$

Short answer

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

Step 1. Given information. 

We have given values of mean $\mu =336$standard deviation $\sigma =6$and$x=325or345$

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

a) Step 2. To find the probability. 

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

$$\begin{gathered} z=\frac{x-\mu }{\sigma }=\frac{325-336}{6}\approx -1.83 \\ z=\frac{x-\mu }{\sigma }=\frac{345-336}{6}=1.50 \end{gathered}$$

Now,

$$\begin{gathered} P(325<X<345)=P(-1.83<X<1.50) \\ =P(Z<1.50)-P(Z<-1.83) \\ =0.9332-0.0336 \\ =0.8996 \\ =89.96\% \end{gathered}$$

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

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

$$\begin{gathered} P(notA)=1-P\left(A\right) \\ P(x<c)=1-P(x\ge c) \\ P(x<c)=1-0.20 \\ P(x<c)=0.80 \end{gathered}$$

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

$$\begin{gathered} z=\frac{x-\mu }{\sigma }=\frac{x-336}{6} \\ \frac{c-336}{6}=0.84 \\ c-336=0.84\times 6 \\ c=336+0.84\left(6\right) \\ c=341.04 \end{gathered}$$