Respuesta :
Let [tex]T[/tex] denote this random variable. Then
[tex]P(T=t)=\begin{cases}\frac1{28}&\text{for }30\le t\le58\\0&\text{otherwise}\end{cases}[/tex]
a.
[tex]P(T>52)=\displaystyle\int_{52}^\infty P(T=t)\,\mathrm dt=\frac1{28}\int_{52}^{58}\mathrm dt=\frac6{28}=\frac3{14}[/tex]
b.
[tex]P(36<T<41)=\displaystyle\int_{36}^{41}P(T=t)\,\mathrm dt=\frac1{28}\int_{36}^{41}\mathrm dt=\frac5{28}[/tex]
c.
[tex]P(T=42.63)=0[/tex]
Using the uniform distribution, it is found that there is a:
a) 0.2143 = 21.43% probability that the student requires more than 52 minutes to complete the quiz.
b) 0.1786 = 17.86% probability that the student completes the quiz in a time between 36 and 41 minutes.
c) 0% probability that the student completes the quiz in exactly 42.63 minutes.
Uniform probability distribution:
An uniform distribution has two bounds, a and b.
- The probability of finding a value between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
- The probability of finding a value above x is:
[tex]P(X > x) = \frac{b - x}{b - a}[/tex]
- The probability of an exact value is 0, that is, P(X = x) = 0.
In this problem, the distribution is uniformly distributed between 30 and 58 minutes, thus: [tex]a = 30, b = 58[/tex].
Item a:
The probability is:
[tex]P(X > 52) = \frac{58 - 52}{58 - 30} = 0.2143[/tex]
0.2143 = 21.43% probability that the student requires more than 52 minutes to complete the quiz.
Item b:
The probability is:
[tex]P(36 \leq X \leq 41) = \frac{41 - 36}{58 - 30} = 0.1786[/tex]
0.1786 = 17.86% probability that the student completes the quiz in a time between 36 and 41 minutes.
Item c:
The probability of an exact value is 0, thus 0% probability that the student completes the quiz in exactly 42.63 minutes.
A similar problem is given at https://brainly.com/question/17088600
