Quesuuns Exercise o.27 Migo (exponential Propaommy Distribution) Consider the

Let's solve the given problem step by step.

a. To find the formula for $P(Z<2)$, we need to integrate the given probability density function from 0 to 2.

$P(Z<2)=\int [0,2]f\left(x\right)dx=\int [0,2]\frac{1}{4}\cdot e^{-\frac{x}{4}}dx$

To solve this integral, we can use u-substitution, where $u=-\frac{x}{4}\text{and}du=-\frac{dx}{4}$.

$P(Z<2)=\int [0,2]\frac{1}{4}\cdot e^{-\frac{x}{4}}dx=-4\int [-\frac{1}{2},0]e^{u}du$

Now, we can solve the integral using the fundamental theorem of calculus:

$P(Z<2)=-4{\left[e^{-u}\right]}_{-\frac{1}{2}}^0=4(e^{\frac{1}{2}}-1)\approx 1.5364$

Therefore, the correct formula for $P(Z<2)$ is not among the options given.

b. To find $P(x<1)$, we need to integrate the probability density function from 0 to 1:

$P(x<1)=\int [0,1]f\left(x\right)dx=\int [0,1]\frac{1}{4}\cdot e^{-\frac{x}{4}}dx$

Again, we can use u-substitution with $u=-\frac{x}{4}\text{and}du=-\frac{dx}{4}$:

$P(x<1)=\int [0,1]\frac{1}{4}\cdot e^{-\frac{x}{4}}dx=-4\int [-\frac{1}{4},0]e^{u}du$

Solving the integral:

$P(x<1)=-4{\left[e^{-u}\right]}_{-\frac{1}{4}}^0=4(1-e^{-\frac{1}{4}})\approx 0.2846$

Therefore, the probability that x is less than 1 is approximately 0.2846.

c. To find $P(2<z<4)$, we need to integrate the probability density function from 2 to 4:

$P(2<z<4)=\int [2,4]f\left(x\right)dx=\int [2,4]\frac{1}{4}\cdot e^{-\frac{x}{4}}dx$

Using u-substitution with $u=-\frac{x}{4}\text{and}du=-\frac{dx}{4}$:

$P(2<z<4)=\int [2,4]\frac{1}{4}\cdot e^{-\frac{x}{4}}dx=-4\int [-1,-\frac{1}{2}]e^{u}du$

Solving the integral:

$P(2<z<4)=-4{\left[e^{-u}\right]}_{-1}^{-\frac{1}{2}}=4(e^{-\frac{1}{2}}-e^{-1})\approx 0.0903$

Therefore, the probability that z is between 2 and 4 is approximately 0.0903.

d. To find $P(a<6)$, we need to substitute $a=\frac{x}{4}$ into the probability density function:

$f\left(a\right)=f\left(\frac{x}{4}\right)=\frac{1}{4}\cdot e^{-\frac{x}{16}}$

Then, we can integrate from 0 to 24:

$P(a<6)=\int [0,24]f\left(a\right)da=\int [0,24]\frac{1}{4}\cdot e^{-\frac{x}{16}}dx$

Using u-substitution with $u=-\frac{x}{16}\text{and}du=-\frac{dx}{16}$:

$P(a<6)=\int [0,24]\frac{1}{4}\cdot e^{-\frac{x}{16}}dx=-16\int [0,-\frac{3}{4}]e^{u}du$

Solving the integral:

$P(a<6)=-16{\left[e^{-u}\right]}_0^{-\frac{3}{4}}=16(1-e^{\frac{3}{4}})\approx 3.7622$

Therefore, the probability that a is less than 6 (or x is less than 24) is approximately 0.3762.