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Quantimethod Act

The document contains two examples of constructing probability distributions for random variables. The first example analyzes student exam results in a class of 100 students and constructs a probability distribution showing the likelihood of students failing 0, 1, 2, or 3 subjects. The second example involves spinning a spinner twice, where landing on numbers 1 and 2 each have a set probability. It constructs a probability distribution showing the likelihoods of obtaining different sum totals from the two spins. Both examples illustrate listing all possible outcomes of a random variable and their associated probabilities to define a probability distribution, which provides useful information for analyzing the performance or expected values of the random variables.

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Kian Lune
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62 views2 pages

Quantimethod Act

The document contains two examples of constructing probability distributions for random variables. The first example analyzes student exam results in a class of 100 students and constructs a probability distribution showing the likelihood of students failing 0, 1, 2, or 3 subjects. The second example involves spinning a spinner twice, where landing on numbers 1 and 2 each have a set probability. It constructs a probability distribution showing the likelihoods of obtaining different sum totals from the two spins. Both examples illustrate listing all possible outcomes of a random variable and their associated probabilities to define a probability distribution, which provides useful information for analyzing the performance or expected values of the random variables.

Uploaded by

Kian Lune
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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1.

In a class of 100 students, 80 students passed in all subjects, 10 failed in one subject, 7
failed in two subjects and 3 failed in three subjects. Find the probability distribution of
the variable for number of subjects a student from the given class has failed in.

P(X = 0) = 80/100 = 0.8 (all students passed in all subjects)

P(X = 1) = 10/100 = 0.1 (10 students failed in one subject)

P(X = 2) = 7/100 = 0.07 (7 students failed in two subjects)

P(X = 3) = 3/100 = 0.03 (3 students failed in three subjects)

Therefore, the probability distribution for X is:

X 0 1 2 3

P(X) 0.8 0.1 0.07 0.03

The probability distribution shows the probabilities of different outcomes of the random variable
X, which is the number of subjects a student from the given class has failed in. The table shows
that the majority of the students (80%) passed in all subjects, while a small percentage of
students failed in one, two, or three subjects. This information can be useful for analyzing the
performance of the students in the class and identifying areas where improvement is needed.

2. The spinner shown below is spun two times. The probability of landing on 1 is 0.25. The
probability of landing on 2 is 0.75. Let X be the sum of the two spins. Construct a
probability distribution for the random variable x.

Let X be the sum of the two spins. We can construct a probability distribution for X by listing all
possible outcomes and their probabilities:

X 2 3 4 5 6 7 8 9

P(X) 0.0625 0.1875 0.1875 0.5625 0 0 0 0

To obtain the probabilities, we use the fact that the probability of a sum being x is the product of
the probabilities of the two spins adding up to x. For example, the probability of X = 3 is the sum
of the probabilities of landing on 1 and 2 (0.25 × 0.75) and landing on 2 and 1 (0.75 × 0.25), which is
0.1875. Similarly, the probability of X = 2 is the product of the probabilities of landing on 1 twice,
which is 0.25 × 0.25 = 0.0625. Since the spinner has only two outcomes, the probabilities for X = 6,
7, 8, and 9 are all zero.

The probability distribution shows the probabilities of different outcomes of the random variable
X, which is the sum of the two spins. The table shows that the most likely outcome is X = 5, which
has a probability of 0.5625. The least likely outcome is X = 2, which has a probability of 0.0625.
This information can be useful for analyzing the expected values and variances of X, as well as for
making decisions based on the outcomes of the two spins.

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