Mathematics: The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable that is uniformly distributed) between 30 and 58 minutes. one student is selected at random. find the probability of the following events. (a) the student requires more than 52 minutes to complete the quiz. (b) the student completes the quiz in a time between 36 and 41 minutes. (c) the student completes the quiz in exactly 42.63 minutes.

The amount of time it takes for a student to complete

Ответ:

turtlesage21

06.01.2022

Mathematics

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:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The probability of finding a value above x is:

$P(X x) = \frac{b - x}{b - a}$

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: a = 30, b = 58 .

Item a:

The probability is:

$P(X 52) = \frac{58 - 52}{58 - 30} = 0.2143$

0.2143 = 21.43% probability that the student requires more than 52 minutes to complete the quiz.

Item b:

The probability is:

$P(36 \leq X \leq 41) = \frac{41 - 36}{58 - 30} = 0.1786$

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 link

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Ответ:

lcyaire

06.01.2022

Mathematics

Let denote this random variable. Then

$P(T=t)=\begin{cases}\frac1{28}&\text{for }30\le t\le58\\0&\text{otherwise}\end{cases}$

$P(T52)=\displaystyle\int_{52}^\infty P(T=t)\,\mathrm dt=\frac1{28}\int_{52}^{58}\mathrm dt=\frac6{28}=\frac3{14}$

P(36

P(T=42.63)=0

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Ответ:

crawford184232323234

04.09.2020

Mathematics

The required area is $\int^{\frac{\pi }{b}}_0 8sin(bt).dt\\$

The appropriate Substitution is $\dfrac{1}{b} \int^\pi _08sinu.du$

Given that,

The area under one arch of the curve y(t) = 8sin(bt) for t ≥ 0 where b is a positive constant.

We have to find,

Set up the definite integral needed to find the area.

Make an appropriate substitution.

According to the question,

The area under one arch of the curve y(t) = 8sin(bt) for t ≥ 0 where b is a positive constant.

The curve y(t) = 8sin(bt) has a period of 2π\b, which is one arch of the curve occur over the interval $0\leq t\leq \frac{\pi }{b}$ .

The area under one arch is given by,

$Area = \int^{\frac{\pi }{b}}_0 8sin(bt).dt\\$

The required area is $Area = \int^{\frac{\pi }{b}}_0 8sin(bt).dt\\$

Appropriate Substitute u= bt ,

Then,

$\dfrac{du}{b} = dt \\\\when \ t=0, \ and \ u=0\\\\when\ t = \dfrac{\pi }{b}, u = \pi$

Then,

The required integral is ,

$\dfrac{1}{b} \int^\pi _08sinu.du$

The appropriate Substitution is $\dfrac{1}{b} \int^\pi _08sinu.du$ .

To know more about Integration click the link given below.

link

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