1/121/61/101723.P(w)1/21/51/54. Find the mean of the probability distribution of the random variable X, which10can take only the values 1, 2, and 3, given that P(1) -, P(2) -, and P(3) -5. The probabilities of a machine manufacturing 0, 1, 2, 3, 4, and 5 defective partsin one day are 0.75, 0.17, 0.04, 0.025, 0.01, and 0.005 respectively. Find the meanof the probability distribution.B. Determine the Variance and Standard Deviation of each random variable. Writeyour answer in your answer sheets.1.1234P(x)1154682.P(z)0.60.10.20.1m13573.P(m)40%25%15%20%4. The random variable X, representing the number of nuts in a chocolate bar hasthe following probability distribution. Compute the variance and standarddeviation.X1245P(x)1332110101010105. The number of items sold per day in a sari-sari store, with its correspondingprobabilities, is shown in the table below. Find the variance and standard deviationof the probability distribution.Number of Items SoldProbabilityP(x)X190.20200.20210.30220.20230.10What I Have LearnedAnswer the following questions in your own understanding.1. How to compute the mean of a discrete random variable? State

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4. Find the mean of the probability distribution of the random variable X, which 10 1 12 can take only the values 1, 2, and 3, given that P(1) P(2) - 33' and P(3) - -- %3D

A First Course in Probability (10th Edition)

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Concept explainers

Contingency Table

A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.

Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

Topic Video

Question

1/12
1/6
1/10
172
3.
P(w)
1/2
1/5
1/5
4. Find the mean of the probability distribution of the random variable X, which
10
can take only the values 1, 2, and 3, given that P(1) -, P(2) -, and P(3) -
5. The probabilities of a machine manufacturing 0, 1, 2, 3, 4, and 5 defective parts
in one day are 0.75, 0.17, 0.04, 0.025, 0.01, and 0.005 respectively. Find the mean
of the probability distribution.
B. Determine the Variance and Standard Deviation of each random variable. Write
your answer in your answer sheets.
1.
1
2
3
4
P(x)
1
1
5
4
6
8
2.
P(z)
0.6
0.1
0.2
0.1
m
1
3
5
7
3.
P(m)
40%
25%
15%
20%
4. The random variable X, representing the number of nuts in a chocolate bar has
the following probability distribution. Compute the variance and standard
deviation.
X
1
2
4
5
P(x)
1
3
3
2
1
10
10
10
10
10
5. The number of items sold per day in a sari-sari store, with its corresponding
probabilities, is shown in the table below. Find the variance and standard deviation
of the probability distribution.
Number of Items Sold
Probability
P(x)
X
19
0.20
20
0.20
21
0.30
22
0.20
23
0.10
What I Have Learned
Answer the following questions in your own understanding.
1. How to compute the mean of a discrete random variable? State

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4. Let X₁ N (0,3²), X₂ N(4,4²), and X3 ~ N(-1,22) be independent random variables. Find P (X₁ + X₂ +3X3 <-8), using the table for standard normal distribution.
23. *An insurance policy pays an individual 100 per day for up to 3 days of hospitalization and 25 per day for each day of hospitalization thereafter. The number of days of hospitalization, X, is a discrete random variable with probability function Pr(X = k) = (1/15)(6 – k), k = 1, 2, 3, 4, 5. Calculate the expected payment for hospitalization under this policy. (213.3)
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