Respuesta :
Answer:
a. [tex]\mathbf{P(Y \leq 60) = 0.3333}[/tex]
b. P(Y>50) = 0.8889
c. E(y) = 67.5 and Standard deviation [tex]\sigma[/tex] = 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 [tex]1\frac{1}{2}[/tex]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;
[tex]1\frac{1}{2}[/tex]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 :
[tex]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[/tex]
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;
[tex]P(Y>50) = \int \limits ^{90}_{50} f(y) dy[/tex]
[tex]P(Y>50) = \int \limits ^{90}_{50} \dfrac {1}{45} dy[/tex]
[tex]P(Y>50) = \dfrac{1}{45}[x]^{90}_{50}[/tex]
[tex]P(Y>50) = (\dfrac{90-50}{45})[/tex]
[tex]P(Y>50) = \dfrac{40}{45}[/tex]
P(Y>50) = 0.8889
Determine the expected painting time and its standard deviation.
Let consider E to be the expected painting time
Then :
[tex]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}[/tex]
[tex]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}[/tex]
To determine the standard deviation, we need to first know what is the value of our variance,
So:
Variance [tex]\sigma^2[/tex] = E(x²) - [E(x)]²
Variance [tex]\sigma^2[/tex] = 4725 - (67.5)²
Variance [tex]\sigma^2[/tex] = 4725 - 4556.25
Variance [tex]\sigma^2[/tex] = 168.75
Standard deviation [tex]\sigma[/tex] = [tex]\sqrt{variance}[/tex]
Standard deviation [tex]\sigma[/tex] = [tex]\sqrt{168.75}[/tex]
Standard deviation [tex]\sigma[/tex] = 12.99
