R2.6 Horse pregnancies Bigger 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 a mean of 336 days and a standard deviation of 3 days. Use the 68-95-99.7 rule to answer the following questions.

(a) Almost all $(99.7\%)$horse pregnancies fall in what range of lengths?

(b) What percent of horse pregnancies are longer than 339 days? Show your work.

Mean $=336days$

Standard deviation$=3days.$

$97\%$horse pregnancies fall in$=?$

Given:

$\mu =336$

$\sigma =3$

About $99.7\%$of all data values are within three standard deviations of the mean according to the $68-95-99.7$rule:

$$\begin{gathered} \mu -3\sigma =336-3\left(3\right) \\ =327 \end{gathered}$$

$$\begin{gathered} \mu +3\sigma =336+3\left(3\right) \\ =345 \end{gathered}$$

Mean$=336$days

Standard deviation $=3$days.

Percent of horse pregnancies are longer than 339 days $=?$

Given:

$\mu =336$

$\sigma =3$

$x=339$

By dividing the standard deviation by the mean of $x$, you get the$z-$score.

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{339-336}{3} \\ =1 \end{gathered}$$

Thus $339$is$1$standard deviation above the mean.

There is a localid="1649335060742" $68\%$chance that the data lies within $1$standard deviation of the data.

Since the distribution is symmetric around $0$, we can then infer that,

localid="1649335067566" $\frac{100\mathrm{\%}-68\mathrm{\%}}{2}=16\mathrm{\%}$

of the data is above$1$standard deviation.