91影视

Mean of Jill's scores$=170$

Standard deviation of Jill's scores$=20$

Jill's scores are approximately normal.

Mean of Jack's scores$=160$

Standard deviation of Jack's scores$=15$

Jack's scores are approximately normal

If the two events are independent and approximately follow a normal distribution

$X-Y~N\left({渭}_x-{渭}_y,{蟽}_x^2+{蟽}_y^2\right)$

Z score

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ where渭isthemean \\ 蟽isthes\tan darddeviation \end{gathered}$$

localid="1647526523734" role="math" $$\begin{gathered} \mathit{\text{闯颈鈭糔}}\left(170,{20}^2\right)approximately \\ Ja\mathit{~}N\left(160,{15}^2\right)approximately \end{gathered}$$

The probability required is

localid="1647526584694" role="math" $$\begin{gathered} \mathit{\text{P}}\text{(}\mathit{\text{Ja>Ji}}\text{)} \\ \mathit{\text{P}}\text{(}\mathit{\text{Ja-Ji}}\text{>0)} \end{gathered}$$

The distribution will be

localid="1647526601842" role="math" $$\begin{gathered} \mathit{\text{Ja-闯颈鈭糔}}\left(160-170,{15}^2+{20}^2\right)approximately \\ Ja\mathit{-}Ji\mathit{}\mathit{~}N\left(-10,{25}^2\right)approximately \end{gathered}$$

find the probability using the continuity correction

localid="1647526838766" role="math" $P(Ja\mathit{-}Ji\mathit{>}\mathit{0})\mathit{=}P(Ja\mathit{-}Ji\mathit{>}0.5)$

localid="1647526873297" $$\begin{gathered} So, \\ P\left(\frac{\mathit{(}\mathit{J}\mathit{a}\mathit{-}\mathit{J}\mathit{i}\mathit{)}\mathit{-}\mathit{渭}}{\mathit{蟽}}>\frac{0.5-(-10)}{25}\right) \\ =P\left(z>0.42\right) \\ =1-P\left(z鈮�0.42\right) \end{gathered}$$

Reading from the table

$$\begin{gathered} P\left(z鈮�0.42\right)=0.6628 \\ P\left(z鈮�0.42\right)=1-0.6628 \\ =0.3372 \end{gathered}$$

If X and Y are independent and approximately follow a normal distribution

$X+Y~N\left({渭}_x+{渭}_y,{蟽}_x^2+{蟽}_y^2\right)approximately$

The distribution will be

$$\begin{gathered} Ja\mathit{+}Ji\mathit{~}N\left(160+170,{15}^2+{20}^2\right)approximately \\ Ja\mathit{+}Ji\mathit{}\mathit{~}N\left(330,{25}^2\right)approximately \end{gathered}$$

the probability required is

$P(Ja+Ji>350)$

using continuity correction

localid="1647526984514" role="math" $$\begin{gathered} P(Ja+Ji>350)=P(Ja+Ji>350.5) \\ P\left(\frac{(Ja+Ji)-渭}{蟽}>\frac{350.5-330)}{25}\right) \\ =P\left(z>0.82\right) \\ =1-P\left(z鈮�0.82\right) \\ Fromztables \\ P\left(z鈮�0.82\right)=0.7939 \\ P\left(z>0.82\right)=1-0.7939 \\ P\left(z>0.82\right)=0.2061 \end{gathered}$$