Q. 6.30 - A First Course in Probability

Question

Jill’s bowling scores are approximately normally distributed with mean $170$ and standard deviation $20$, while Jack’s scores are approximately normally distributed with mean $160$ and standard deviation $15$. If Jack and Jill each bowl one game, then assuming that their scores are independent random variables, approximate the probability that

(a) Jack’s score is higher; 

(b) the total of their scores is above $350$ 

Step-by-Step Solution

Answer

(a) Probability that Jack’s score is higher than Jill is $0.3372$

(b) Probability that the total of their scores is above $350$ is $0.2061$

Step 1: Given information (part a)

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

Step 2: Explanation (part a)

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

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

Z score

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ where \mu is the mean \\ \sigma is the s\tan dard deviation \end{gathered}$$

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

The probability required is

$$\begin{gathered} \mathit{\text{P}}\text{(}\mathit{\text{Ja >Ji}}\text{)} \\ \mathit{\text{P}}\text{(}\mathit{\text{Ja-Ji}}\text{ > 0)} \end{gathered}$$

The distribution will be

$$\begin{gathered} \mathit{\text{Ja-Ji ∼N}}\left(160-170,{15}^2+{20}^2\right) approximately \\ Ja\mathit{-}Ji\mathit{ }\mathit{~}N\left(-10,{25}^2\right) approximately \end{gathered}$$

find the probability using the continuity correction 

$P(Ja\mathit{-}Ji\mathit{>}\mathit{0})\mathit{=}P(Ja\mathit{-}Ji\mathit{>}0.5)$

$$\begin{gathered} So, \\ P\left(\frac{\mathit{(}\mathit{J}\mathit{a}\mathit{-}\mathit{J}\mathit{i}\mathit{)}\mathit{-}\mathit{\mu }}{\mathit{\sigma }}>\frac{0.5-(-10)}{25}\right) \\ =P\left(z>0.42\right) \\ =1-P\left(z\le 0.42\right) \end{gathered}$$

Reading from the table

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

Step 1: Given information (part b)

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

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

Step 2: Explanation (part b)

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

$P(J a+J i>350)$

using continuity correction

$$\begin{gathered} P(Ja+Ji>350)=P(Ja+Ji>350.5) \\ P\left(\frac{(Ja+Ji)-\mu }{\sigma }>\frac{350.5-330)}{25}\right) \\ =P\left(z>0.82\right) \\ =1-P\left(z\le 0.82\right) \\ From z tables \\ P\left(z\le 0.82\right)=0.7939 \\ P\left(z>0.82\right)=1-0.7939 \\ P\left(z>0.82\right)=0.2061 \end{gathered}$$