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Setler79 [48]
2 years ago
8

The following table shows the probability distribution for a discrete random variable. X 12 15 17 20 22 24 25 30 P(X) 0.07 0.21

0.17 0.25 0.05 0.04 0.13 0.08 What is the mean of this discrete random variable? (That is, what is E(X), the expected value of X?) A. 20 B. 19.59 C. 20.625 D. 21

Mathematics
2 answers:
Andrews [41]2 years ago
8 0

The mean of this discrete random variable is B. 19.59

<h3>Further explanation</h3>

The probability of an event is defined as the possibility of an event occurring against sample space.

Let us tackle the problem !

<em>The mean of </em><em>discrete random variable</em><em> could be calculated by using this following formula.</em>

\large {\boxed {E(x) = \mu = \sum x_ip_i} }

\texttt{ }

<u>Given:</u> <em>The probability distribution for a discrete random variable:</em>

\boxed{\texttt{    X\ \      }}\boxed{\texttt{ 12\  \ \ }}\boxed{\texttt{ 15\  \ \ }}\boxed{\texttt{ 17\  \ \ }}\boxed{\texttt{ 20\  \ \ }}\boxed{\texttt{ 22\  \ \ }}\boxed{\texttt{ 24\  \ \ }}\boxed{\texttt{ 25\  \ \ }}\boxed{\texttt{ 30\  \ \ }}

\boxed{\texttt{P(X)}}\boxed{\texttt{ 0.07 }}\boxed{\texttt{ 0.21 }}\boxed{\texttt{ 0.17 }}\boxed{\texttt{ 0.25 }}\boxed{\texttt{ 0.05 }}\boxed{\texttt{ 0.04 }}\boxed{\texttt{ 0.13 }}\boxed{\texttt{ 0.08 }}

<em>From the above table, the mean can be calculated:</em>

E(x) = \mu = 12 ( 0.07 ) + 15( 0.21 ) + 17(0.17) + 20( 0.25 ) + 22(0.05) + 24(0.04) + 25(0.13) + 30(0.08)

E(x) = \mu = 0.84 + 3.15 + 2.89 + 5 + 1.1 + 0.96 + 3.25 + 2.4

E(x) = \mu = 19.59

<h2><u>Conclusion:</u></h2>

The mean of this discrete random variable is 19.59

<h3>Learn more</h3>
  • Different Birthdays : brainly.com/question/7567074
  • Dependent or Independent Events : brainly.com/question/12029535
  • Mutually exclusive : brainly.com/question/3464581

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Probability

Keywords: Probability , Sample , Space , Six , Dice , Die , Binomial , Distribution , Mean , Variance , Standard Deviation , DIscrete , Random , Variable , Expected , Value , E(x) , Table

Volgvan2 years ago
7 0

The mean of this discrete random variable can be calculated by finding for the summation of the weighted average:

Mean = Summation (Xi * P(Xi))

Therefore,

Mean = 12 (0.07) + 15 (0.21) + 17 (0.17) + 20 (0.25) + 22 (0.05) + 24 (0.04) + 25 (0.13) + 30 (0.08)

Mean = 19.59

<span>Therefore the answer is letter B. 19.59</span>

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What is the error due to using linear interpolation to estimate the value of sinxsin⁡x at x = \pi/3? your answer should have at
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<h3>Answer:</h3>
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  • using y = (x -π/4 +1)/√2, the error is about 0.02620
<h3>Step-by-step explanation:</h3>

The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.

If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...

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... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812

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If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...

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6 0
2 years ago
6.5% of people in the U.S. Have A- blood type. You randomly select 6 Americans and ask them if their blood type is A-. a) Find t
Keith_Richards [23]

Answer:

a) 7.54189*10^-8

b) 0.99999999

c) E ( X ) = 0.39 , s ( X ) = 0.604

Step-by-step explanation:

Solution:-

- We will assume the proportion of people with A- blood group is independent and remains constant for a fairly small sample of n = 6 Americans selected at random.

- We will denote a random variable X = The number of americans out of 6 that have blood group type A-.

- The random variable is assumed to follow binomial distribution.

- The probability of success is the proportion of people in U.S that have blood group type A-, p = 0.065:

                        X ~ Bin ( 6 , 0.065 )

Where, r represents the number of americans out of selected 6 have blood group A-. The pmf of a binomial variate is given as:

                       P ( X = r ) = nCr * ( p ) ^r * ( 1 - p ) ^ ( n - r )

a) Find the probability that all 6 are type A-

- We will pmf given above and set r = 6. And evaluate the resulting probability:

                     P ( X = 6 ) = 6C6 * ( p )^6 * ( 1  - p ) ^ ( 0 )

                                      = p^6

                                      = ( 0.065 )^6  

                                      = 7.54189*10^-8

b) Find the probability that at most 4 of them are type A-

- We will pmf given above and evaluate the following expression:

                     P ( X ≤ 4 ) = 1 - P ( X = 5 ) - P ( X = 6 )

                     P ( X ≤ 4 ) = 1 - 6C5 * ( p )^5 * ( 1  - p ) ^ ( 1 ) - p^6

                                      = 1 - 6*(0.065)^5 ( 0.935 ) - 0.065^6

                                      = 1 - 6.50923*10^-8 - 7.54189*10^-8        

                                      = 0.99999999          

c) Find the mean and standard deviation.

- The mean E ( X ) of the defined random variable distributed binomially is given by:

                     E ( X ) = n*p = 6*0.065 = 0.39 people

- The mean s( X ) of the defined random variable distributed binomially is given by:

   

                    s ( X ) = √n*p*q = √(6*0.065*0.935) = 0.604

5 0
2 years ago
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