Plz plz plz help due in tommorw!!!!$

In answering a question on a multiple-choice test, a student either knows the correct answer or guesses it. Let 0.7 be the proba

Dahasolnce [82]

Answer:

The probability is 0.9211

Step-by-step explanation:

Let's call K the event that the student know the answer, G the event that the student guess the answer and C the event that the answer is correct.

So, the probability P(K/C) that a student knows the answer to a question, given that she answered it correctly is:

P(K/C)=P(K∩C)/P(C)

Where P(C) = P(K∩C) + P(G∩C)

Then, the probability P(K∩C) that the student know the answer and it is correct is:

P(K∩C) = 0.7

On the other hand, the probability P(G∩C) that the student guess the answer and it is correct is:

P(G∩C) = 0.3*0.2 = 0.06

Because, 0.3 is the probability that the student guess the answer and 0.2 is the probability that the answer is correct given that the student guess the answer.

Therefore, The probability P(C) that the answer is correct is:

P(C) = 0.7 + 0.06 = 0.76

Finally, P(K/C) is:

P(K/C) = 0.7/0.76 = 0.9211

7 0

2 years ago

Select the correct answer from each drop-down menu.

sergij07 [2.7K]

Answer:

7.8 and 18.8 is your answer

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

At the start of the month, the value of an investment was $73.42. By the end of the month, the value of the investment changed b

Aleonysh [2.5K]

Answer:

0.8%.

Step-by-step explanation:

Given:

At the start of the month, the value of an investment was $73.42.

By the end of the month, the value of the investment changed by a loss of $13.53.

Question asked:

What was the value, in dollars, of the investment at the end of the month?

What was the percent loss?

Solution:

At the start of the month, the value of an investment = $73.42.

Loss = $13.53.

The value, of the investment at the end of the month = The value of an investment at the start of the month - loss

The value, of the investment at the end of the month = $73.42 - $13.53

= $59.89

Thus, the value, of the investment at the end of the month is $59.89.

Now, we will find percent loss:-

$Percent\ loss = \frac{Loss}{value\ of \ investment\ at \ start}$

$=\frac{59.89}{73.42} \\=0.81\%$

Thus, percent loss on investment by the end of the month is 0.8%.

7 0

2 years ago

56:12

Gre4nikov [31]

Answer:

The official answer is {C} ive previously taken the test

Step-by-step explanation:

I have taken the test before hand.

5 0

1 year ago

Test scores of the student in a school are normally distributed mean 85 standard deviation 3 points. What's the probability that

Mrrafil [7]

Answer:

The probability that a random selected student score is greater than 76 is $\\ P(x>76) = 0.99865$ .

Step-by-step explanation:

The Normally distributed data are described by the normal distribution. This distribution is determined by two parameters, the population mean $\\ \mu$ and the population standard deviation $\\ \sigma$ .

To determine probabilities for the normal distribution, we can use the standard normal distribution, whose parameters' values are $\\ \mu = 0$ and $\\ \sigma = 1$ . However, we need to "transform" the raw score, in this case x = 76, to a z-score. To achieve this we use the next formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

And for the latter, we have all the required information to obtain z. With this, we obtain a value that represent the distance from the population mean in standard deviations units.

<h3>The probability that a randomly selected student score is greater than 76</h3>

To obtain this probability, we can proceed as follows:

First: obtain the z-score for the raw score x = 76.

We know that:

$\\ \mu = 85$

$\\ \sigma = 3$

$\\ x = 76$

From equation [1], we have:

$\\ z = \frac{76 - 85}{3}$

Then

$\\ z = \frac{-9}{3}$

$\\ z = -3$

Second: Interpretation of the previous result.

In this case, the value is three (3) standard deviations below the population mean. In other words, the standard value for x = 76 is z = -3. So, we need to find P(x>76) or P(x>-3).

With this value of $\\ z = -3$ , we can obtain this probability consulting the cumulative standard normal distribution, available in any Statistics book or on the internet.

Third: Determination of the probability P(x>76) or P(x>-3).

Most of the time, the values for the cumulative standard normal distribution are for positive values of z. Fortunately, since the normal distributions are symmetrical, we can find the probability of a negative z having into account that (for this case):

$\\ P(z>-3) = 1 - P(z>3) = P(z$

Then

Consulting a cumulative standard normal table, we have that the cumulative probability for a value below than three (3) standard deviations is:

$\\ P(z$

Thus, "the probability that a random selected student score is greater than 76" for this case (that is, $\\ \mu = 85$ and $\\ \sigma = 3$ ) is $\\ P(x>76) = P(z>-3) = P(z$ .

As a conclusion, more than 99.865% of the values of this distribution are above (greater than) x = 76.

We can see below a graph showing this probability.

As a complement note, we can also say that:

$\\ P(z3)$

Which is the case for the probability below z = -3 [P(z<-3)], a very low probability (and a very small area at the left of the distribution).

5 0

2 years ago