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Vikentia [17]
3 years ago
8

On a quiz there are four multiple-choice questions worth 3 points each and two true/false questionsworth 1 point each . Each mul

tiple-choice question has five possible choices . If a student randomlyguesses on each question , what is the expected value of the student's score on the test ?
Mathematics
1 answer:
Anastaziya [24]3 years ago
7 0

Answer:

The expected value of the student's score on the test is 3.4.

Step-by-step explanation:

Multiple choice questions:

worth 3 points.

Five possible choices, one of which is correct. So the expected value of the student in each multiple choice question is:

m = 3\frac{1}{5} = 0.6

True/false questions:

worth 1 point.

2 options, one of which is correct, so the expected value for each is:

t = 1\frac{1}{2} = 0.5

What is the expected value of the student's score on the test ?

4 multiple choice, two true-false. So

E = 4m + 2f = 4(0.6) + 2(0.5) = 2.4 + 1 = 3.4

The expected value of the student's score on the test is 3.4.

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(b) The probability that you will have to wait more than 2 standard deviations is 0.4227.

Step-by-step explanation:

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The random variable <em>X</em> is uniformly distributed with parameters <em>a</em> = 0 to <em>b</em> = 15.

The probability density function of <em>X</em> is given as follows:

f_{X}(x)=\frac{1}{b-a};\ a

(a)

The standard deviation of a Uniformly distributed random variable is given by:

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Compute the standard deviation of the random variable <em>X</em> as follows:

SD=\sqrt{\frac{(b-a)^{2}}{12}}

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      =\sqrt{18.75}\\=4.33

Thus, the standard deviation of your waiting time is 4.33 minutes.

(b)

The value representing 2 standard deviations is:

X=2\times SD=2\times4.33=8.66

Compute the value of P (X > 8.66) as follows:

P(X>8.66)=\int\limits^{10}_{8.66} {\frac{1}{15-0}}\, dx\\

                    =\frac{1}{15}\times \int\limits^{10}_{8.66} {1}\, dx\\

                    =\frac{1}{15}\times |x|^{15}_{8.66}\\

                    =\frac{15-8.66}{15}\\

                    =0.4227

Thus, the probability that you will have to wait more than 2 standard deviations is 0.4227.

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