91影视

Mean $=336days$

Standard deviation$=3days.$

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

Given:

$渭=336$

$蟽=3$

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

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

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

Mean$=336$days

Standard deviation $=3$days.

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

Given:

$渭=336$

$蟽=3$

$x=339$

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

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

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" $\frac{100\mathrm{\%}鈭�68\mathrm{\%}}{2}=16\mathrm{\%}$

of the data is above$1$standard deviation.