The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let X = the time needed to change the oil on a car.

a. Write the random variable X in words. X = __________________.

b. Write the distribution.

c. Graph the distribution.

d. Find P (x > 19).

e. Find the 50th percentile.

Given in the question that, the amount of time a service technician needs to change the oil in a car is uniformly distributed between $11$and $21$minutes.

We need to write the random variable $X$in words.

According to the information from the question, the amount of time is uniformly distributed with parameter $(11,21)$.

Therefore, the random variable $X$can be described as

$X=Amountoftimeneededtochangetheoil$

Given in the question that, the amount of time a service technician needs to change the oil in a car is uniformly distributed between $11$and $21$minutes.

We have to write the distribution.

From the information given in the question, they clearly mentioned that the amount of time is uniformly distributed.

Therefore, the distribution is uniform distribution.

So,$X~U(11,21)$

According to the information, the variable $a=11$and $b=21$.

We have to graph the distribution.

Let's find the probability density function first,

The height of the data will be

$f\left(x\right)=\frac{1}{b-a}$

$=\frac{1}{21-11}$

$=\frac{1}{10}$

The graph of the distribution given below,

From the information,

$a=11$

$b=21$

We have to find$P(x>19)$

The required probability can be computed as below,

$P(x>19)=\text{base}\times \text{height}$

$=(21-19)\times \left(\frac{1}{10}\right)$

localid="1648011304762" $=2\times \frac{1}{10}$

$=0.2$

From the information,

$a=11$

$b=21$

We have to find the${50}^{th}$percentile.

Let's calculate the ${50}^{th}$by using the given formula,

$P(x<k)=\text{base}\times \text{height}$

$0.50=(k-11)\times \left(\frac{1}{10}\right)$

$k-11=0.50\times 10$

$k=5+11$

$=16$

Hence, The value of the ${50}^{th}$percentile is $16$