Question

The mean of $T$is

(a)$110$

(b)$140$

(c)$180$

(d)$195$

(e)$250$

Short answer

The mean of$T$is option (c)$180.$

Step-by-step solution

Concept Introduction 

A mean is a measurement of center of gravity in a probability distribution, along with mean, median, and mode. In statistics, it is a measure of central tendency.

Explanation 

Given:

${\mu }_X=110$

${\sigma }_X=10$

${\mu }_Y=140$

${\sigma }_Y=12$

Property mean and variance (if $X$and $Y$are independent):

localid="1649859475011" $$\begin{gathered} {\mu }_{aX+bY}=\left(a{\mu }_X+b{\mu }_Y\right) \\ {\sigma }_{aX+bY}^2=\left(a^2{\mu }_X^2+b^2{\mu }_Y^2\right) \end{gathered}$$

The obtained results are:

localid="1649859480196" $$\begin{gathered} {\mu }_{X+Y/2}={\mu }_X+\frac{1}{2}{\mu }_Y \\ =110+\frac{1}{2}\left(140\right) \\ =180 \end{gathered}$$

localid="1649859485292" $$\begin{gathered} {\sigma }_{X+Y/2}=\sqrt{{\sigma }_X^2+\frac{1}{2^2}{\sigma }_{X_2}^2} \\ =\sqrt{\left({10}^2+\frac{1}{4}{12}^2\right)} \\ \approx 11.66 \end{gathered}$$