Question

The body paint, an automobile body paint shop, has determined that the painting time of automobiles is uniformly distributed and that the required time ranges between 45 minutes to 11/2hours.
What is the probability that the painting time will be less than or equal to an hour?

What is the probability that the painting time will be more than 50 minutes?

Determine the expected painting time and its standard deviation.

Answer 1

Answer:

a. $\mathbf{P(Y \leq 60) = 0.3333}$

b. P(Y>50) = 0.8889

c. E(y) = 67.5 and  Standard deviation $\sigma$ = 12.99

Step-by-step explanation:

From the information given :

an automobile body paint shop, has determined that the painting time of automobiles is uniformly distributed and that the required time ranges between 45 minutes to $1\frac{1}{2}$hours.

The objective is to determine  the probability that the painting time will be less than or equal to an hour?

since 60 minutes make an hour;

$1\frac{1}{2}$hours = 60 +30 minutes = 90 minutes

Let Y be the painting time of the automobile; then,

the probability that  the painting time will be less than or equal to an hour ca be  computed as :

$P(Y \leq 60) = \int ^{60}_{45} f(y) dy \\ \\ \\ P(Y \leq 60) = \int ^{60}_{45} \dfrac{1}{45} dy \\ \\ \\ P(Y \leq 60) = \dfrac{1}{45} \begin {pmatrix} x\end {pmatrix}^{60}_{45} \\ \\ \\ P(Y \leq 60) = \dfrac{60-45}{45 } \\ \\ \\ P(Y \leq 60) = \dfrac{15}{45} \\ \\ \\ P(Y \leq 60) = \dfrac{1}{3} \\ \\ \\ P(Y \leq 60) = 0.3333$

What is the probability that the painting time will be more than 50 minutes?

The probability that the painting will be more than 50 minutes is P(Y>50)

So;

$P(Y>50) = \int \limits ^{90}_{50} f(y) dy$

$P(Y>50) = \int \limits ^{90}_{50} \dfrac {1}{45} dy$

$P(Y>50) = \dfrac{1}{45}[x]^{90}_{50}$

$P(Y>50) = (\dfrac{90-50}{45})$

$P(Y>50) = \dfrac{40}{45}$

P(Y>50) = 0.8889

Determine the expected painting time and its standard deviation.

Let consider E to be the expected painting time

Then :

$E(y) = \int \limits ^{90}_{45} y f(y) dy \\ \\ \\ E(y) = \int \limits ^{90}_{45} y \dfrac{1}{45} dy \\ \\ \\ E(y) = \dfrac{1}{45} [\dfrac{y^2}{2}]^{90}_{45} \\ \\ \\ E(y) = \dfrac{1}{45}[\dfrac{(90^2-45^2)}{2}] \\ \\ \\ E(y) = \dfrac{1}{45} (\dfrac{6075}{2}) \\ \\ \\ E(y) = \dfrac{1}{45} \times 3037.8 \\ \\ \\ \mathbf{E(y) = 67.5}$

$E(y^2) = \int \limits ^{90}_{45} y^2 f(y) dy \\ \\ \\ E(y^2) = \int \limits ^{90}_{45} y^2 \dfrac{1}{45} dy \\ \\ \\ E(y^2) = \dfrac{1}{45} [\dfrac{y^3}{3}]^{90}_{45} \\ \\ \\ E(y^2) = \dfrac{1}{45}[\dfrac{(90^3-45^3)}{3}] \\ \\ \\ E(y^2) = \dfrac{1}{45} (\dfrac{637875}{3}) \\ \\ \\ E(y^2) = \dfrac{1}{45} \times 2126.25 \\ \\ \\ \mathbf{E(y^2) = 4725}$

To determine the standard deviation, we need to first know what is the value of our variance,

So:

Variance $\sigma^2$ = E(x²) - [E(x)]²

Variance $\sigma^2$ = 4725 - (67.5)²

Variance $\sigma^2$ = 4725 - 4556.25

Variance $\sigma^2$ = 168.75

Standard deviation $\sigma$ = $\sqrt{variance}$

Standard deviation $\sigma$ = $\sqrt{168.75}$

Standard deviation $\sigma$ = 12.99