What is the value of x?

The amount of time it takes for a student to complete a statistics quiz is uniformly distributed (or, given by a random variable

topjm [15]

Answer:

(A) 0.15625

(B) 0.1875

(C) Can't be computed

Step-by-step explanation:

We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let X = Amount of time taken by student to complete a statistics quiz

So, X ~ U(32 , 64)

The PDF of uniform distribution is given by;

f(X) = $\frac{1}{b-a}$ , a < X < b where a = 32 and b = 64

The CDF of Uniform distribution is P(X <= x) = $\frac{x-a}{b-a}$

(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)

P(X > 59) = 1 - P(X <= 59) = 1 - $\frac{x-a}{b-a}$ = 1 - $\frac{59-32}{64-32}$ = $1-\frac{27}{32}$ = 0.15625

(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)

P(X <= 43) = $\frac{43-32}{64-32}$ = $\frac{11}{32}$ = 0.34375

P(X < 37) = $\frac{37-32}{64-32}$ = $\frac{5}{32}$ = 0.15625

P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

(C) Probability that student complete the quiz in exactly 44.74 minutes

= P(X = 44.74)

The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.

3 0

2 years ago

Which of the following is the graph of y = negative StartRoot x EndRoot + 1? On a coordinate plane, an absolute value curve curv

mestny [16]

Answer:

B. On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

Step-by-step explanation:

<u>Graph of the function:</u>

y = -√x + 1

The domain is x ≥ 0, the range y ≤ 1

Correct answer choice is B

On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

<em>The graph is attached</em>

7 0

2 years ago

Read 2 more answers

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

$P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}$

And if we want to find the joint probability we just need to do this:

$P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)$

And if we multiply the probabilities founded we got:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

8 0

2 years ago

Nar donated to 4 charities last year. She gave £175 to each of these charities. This year Rana wants to donate the same total am

LekaFEV [45]

Answer:

£116.67

Step-by-step explanation:

Given;

Nar donated to 4 charities last year. She gave £175 to each of these charities.

The total amount Nar donated is;

Nar = 4 × £175 = £700

For Rana to donate the same amount equally among 6 charity, to determine the amount each charity would receive this year, we need to divide the total amount by the number of charities;

A = £700/6

A = £116.67

7 0

2 years ago

What is the product? StartFraction x squared minus 16 Over 2 x + 8 EndFraction times StartFraction x cubed minus 2 x squared + x

zubka84 [21]

Answer:

[x(x - 1)(x - 4)]/(2(x + 4))

Step-by-step explanation:

We want to find;

[(x² - 16)/(2x + 8)] * [(x³ - 2x² + x)/(x² + 3x - 4)]

Now,

x² - 16 can be factorized as;

(x + 4)(x - 4)

Also, 2x + 8 can be factorized as;

2(x + 4)

Also, (x³ - 2x² + x) can factorized as;

x[x² - 2x + 1] = x[(x - 1)(x - 1)]

Also,(x² + 3x - 4) can be factorized out as; (x - 1)(x + 4)

So plugging in these factorized forms into the equation in the question, we have;

[(x + 4)(x - 4)/(2(x + 4))] * [x[(x - 1)(x - 1)] /((x - 1)(x + 4))

This gives;

((x - 4)/2) * x(x - 1)/(x +4)

This gives;

[x(x - 1)(x - 4)]/(2(x + 4))

8 0

2 years ago

Read 2 more answers