Chapter 6, Problem T6.13SPT

(a)

To determine

To compute: Mean and standard deviation of D

Answer to Problem T6.13SPT

Mean and standard deviation of D are 25 minutes and 11.180 minutes respectively.

Explanation of Solution

Given information:

E is the time Ed takes to do his maths homework.

A is the time Adelaide takes to do maths her homework.

Mean time of E $=\text{ 25 minutes}$

Standard deviation of E $=\text{ 5 minutes}$

Mean time of A $=\text{ 5}0\text{ minutes}$

Standard deviation of A $=\text{ 1}0\text{ minutes}$

Both E and A are independent of each other.

Concept used:

If two variables are independent,

  $\text{Mean }\left(\text{X }-\text{ Y}\right)=\text{ E }\left(\text{X}\right)–\text{ E }\left(\text{Y}\right)$

  $\text{Var }\left(\text{X }–\text{ Y}\right)=\text{ Var }\left(\text{X}\right)+\text{ Var }\left(\text{Y}\right)$

Calculation:

  $\text{D }=\text{ A }–\text{ E}$

Mean of D

  $\text{E }\left(\text{D}\right)=\text{ E }\left(\text{A }–\text{ E}\right)$

  $=\text{ E }\left(\text{A}\right)–\text{ E }\left(\text{E}\right)$

  $=\text{ 5}0–\text{ 25 }=\text{ 25 minutes}$

Variance of D

  $\text{V }\left(\text{A }–\text{ E}\right)=\text{ V }\left(\text{A}\right)+\text{ V }\left(\text{E}\right)$

  $=\text{ 1}{0}^{\text{2}}+{\text{ 5}}^{\text{2}}$

  $=\text{ 125 }$

Standard deviation D $=\sqrt{\text{V (D)}}=\sqrt{125}=11.180\text{ minutes}$

So,

  $\text{D }~\text{ N }\left(\text{25},\text{ 125}\right)$

(b)

To determine

To compute: Probability that Ed spent more time than Adelaide.

Answer to Problem T6.13SPT

Probability that Ed spent more time than Adelaide is 0.987322.

Explanation of Solution

Calculation:

  $\text{P }\left(\text{E }>\text{ A}\right)=\text{ P }\left(\text{E }–\text{ A }>0\right)$

  $\text{P }\left(\text{D }>0\right)=\text{ P }\left(\frac{D-{{\displaystyle \mu }}_D}{{{\displaystyle \sigma }}_D}>\frac{0-25}{11.180}\right)$

  $\text{P }\left(\text{z }>-\text{2}.\text{236}\right)=\text{ P }\left(\text{z }<\text{ 2}.\text{236}\right)$

From z tables,

  $\text{P }\left(\text{z }<\text{ 2}.\text{23}\right)=0.\text{98713}$

  $\text{P }\left(\text{z }<\text{ 2}.\text{24}\right)=0.\text{98745}$

  $\text{P }\left(\text{z }<\text{ 2}.\text{236}\right)=0.98713+(0.98745-0.98713)\times 0.6$

Therefore,

  $\text{P }\left(\text{z }<\text{ 2}.\text{236}\right)=0.\text{987322}$