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The time (in years) after reaching age $60$that it takes an individual to retire is approximately exponentially distributed with a mean of about five years.

The definition of the random variable X is given as the time after $60$years of age to retirement.

The time (in years) after reaching age $60$that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Therefore, $X$is continuous.

According to the provided information on the problem, the random variable $X$is exponentially distributed and can be defined as below:

localid="1651916931002" $$\begin{gathered} X~Exp\left(\frac{1}{渭}\right) \\ X~Exp\left(\frac{1}{5}\right) \\ X~Exp\left(0.2\right)where,m=0.2 \end{gathered}$$

The mean can be represented as below:

$渭=5$

The standard deviation is represented as follows:

$蟽=渭=5$

It is given that $X~Exp(0.2)$so it clear that,

$m=0.2$
And, the general form of the probability density function of the exponential distribution is given below,

$f\left(x\right)=m{e}^{-mx}=f\left(x\right)=(0.2)e^{-(0.2)x}$

The maximum value of $f\left(x\right)$which will lie on the y-axis and at $x=0$will be:

$f\left(x\right)=(0.2)e^{-(0.2脳0)}=0.2$

The value of $f\left(x\right)$for different values of $x$, we get

From the above table, the graph of the probability distribution is given as:

The probability distribution of an Exponential distribution is:

$f\left(x\right)=位e^{鈭�位x};x>0$

person retired after age $70$years.

Then the number of years after the retirement age is$70-60=10$years.

The probability of $(X>10)$is as follows:

${鈭�}_{10}^{\mathrm{鈭�}}鈥�位e^{鈭�位x}dx=位{\left|\frac{e^{鈭�位x}}{鈭�位}\right|}_{10}^{\mathrm{鈭�}}=e^{鈭�0.20脳10}鈭�0=0.1353$

According to the given information, the retirement age is $60$years. So, the number of people retire before the age$65.$

The probability distribution of an Exponential distribution is:

$f\left(x\right)=位e^{鈭�位x};x>0$

The person retired after age $80$years.

Then the number of years after the retirement age is, $80-60=20$years.

The probability of localid="1651510500745" $X>20$is as follows:

$P={鈭�}_{20}^{\mathrm{鈭�}}鈥�位e^{鈭�位x}dx=位{\left|\frac{e^{鈭�位x}}{鈭�位}\right|}_{20}^{\mathrm{鈭�}}=e^{鈭�0.20脳20}鈭�0=0.0183$

Total people in a room $n=1000$

The number of people that will not retire yet is given by,

$nP=1000脳0.0183=18.3$