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The percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of$9.848.$

The random variable$X$an be defined as the per cent of persons (ages five and older) in each state who speak a language at home other than English.

Continous as the random variable $X$is exponentially distributed.

The random variable $X$is exponentially distributed and can be defined as below:

$$\begin{gathered} =X~\mathrm{Exp}鈦�\left(\frac{1}{9.848}\right) \\ =X~\mathrm{Exp}鈦�\left(m\right) \\ =X~\mathrm{Exp}鈦�\left(0.1015\right) \end{gathered}$$

The mean can be represented as$渭=9.848$

The standard deviation can be calculated as-

$蟽=渭=9.848$

Given that $X~Exp(0.1015)$so$m=0.1015$
And, the general form of the probability density function of the exponential distribution is given below,

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{me}}^{鈭�\mathrm{mx}}鈬�\mathrm{f}\left(\mathrm{x}\right)=0.1015{\mathrm{e}}^{鈭�0.1015\mathrm{x}}$

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

$f\left(x\right)=(0.1015)e^{-(0.1015)\left(0\right)}=0.1015$

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:

It is known that the cumulative distribution function of the exponential distribution is,

$p(X<x)=1-e^{-mx}$

The probability less than the per cent is less than $12$is,

localid="1651432803210" $$\begin{gathered} p(X<12)=1-e^{-(0.1015脳12)} \\ p(X<12)=1-0.2958 \\ p(X<12)=0.7042 \end{gathered}$$

The required probability can be calculated as,

$$\begin{gathered} P(14<x<8)=P(X<14)-P(X<8) \\ P(14<x<8)=(1-e^{-(0.1015脳14)})-(1-e^{-(0.1015脳8)}) \\ P(14<x<8)=0.4440-0.2415 \\ P(14<x<8)=0.2025 \end{gathered}$$

i. This is due to the bigger value of standard deviation which is present in the collected data.
ii. Due to the huge chance that each state will have English speakers higher and lower than the mean percentage the given percentage is higher than$9.848$