Question 1 Use rules for Discrete Probability Distribution t

Question 1

Use rules for Discrete Probability Distribution
to find missing value in this table:

0.30

0.25

0.20

0.15

1 points

Question 2

Find Expected Value for the random variable
with the following probability distribution:

0.8

1.6

2.1

3.0

1 points

Question 3

Use Binomial Distribution (formula or Appendix Table)
to find probability that you flip the coin 6 times
and will obtain Head exactly 4 times.

Appendix Table for Binomial Distribution

0.500       

0.234

0.356

0.412

1 points

Question 4

Use Appendix Tables for Binomial Distribution
to find probability that with n = 10 and p = 0.2
x will be less than 3: P(x < 3) (x = 0,1,2)

Appendix Table for Binomial Distribution

0.245

0.348

0.677

0.724

1 points

Question 5

Use Appendix Tables for Binomial Distribution
to find probability that with n = 6 and p = 0.2
x will be at least 3: P(x ? 3) (x = 3,4,5,6).

Appendix Table for Binomial Distribution

0.464

0.351

0.284

0.099

x	0	1	2	3	4
P(x)	0.15	0.25	...	0.20	0.10

Solution

(1) Since sum of f(x)=1, 1-0.15-0.25-0.2-0.1=0.3

Answer: 0.3

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(2) Expected value= sum of x*f(x)

=0*0.1+1*0.2+2*0.3+3*0.3+4*0.1

=2.1

Answer: 2.1

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(3) P(Head) = 0.5

Given X~Binomial(n=6, p=0.5)

P(X=x)=6Cx*(0.5^6)

So the probability is

P(X=4) =6C*(0.5^6) = 0.234375

Answer: 0.234

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(4) Given X~Binomial(n=10, p = 0.2)

P(X=x)=10Cx*(0.2^x)*(0.8^(10-x))

So P(x < 3) = P(X=0)+P(X=1)+P(X=2)

=10C0*(0.2^0)*(0.8^(10-0))+..+10C2*(0.2^2)*(0.8^(10-2))

=0.6777995

Answer: 0.677

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(5) Given X~Binomial(n=6, p = 0.2)

P(X=x)=6Cx*(0.2^x)*(0.8^(6-x))

So P(X>=3)= P(X=3)+P(X=4)+P(X=5)+P(X=6)

=6C3*(0.2^3)*(0.8^(6-3))+...+6C6*(0.2^6)*(0.8^(6-6))

=0.09888

Answer: 0.099