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section 16 probability distributions and expected values activity

Section 16 Probability Distributions and Expected Values Activity 2 The following are probability distributions for variable
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Answer #1

Solution:

1. We have to find the P(x=20)

We know that the sum of probabilities is 1. Therefore, we have:

Σ P(x) = 1

0.1 +0.3+0.1 +0.2 +0.2+ P(x = 20) = 1

P(x = 20) = 1-0.9

P(x = 20) = 0.1

a.

P(x < 9) = P(x = 2) + P(x = 3) + P(x = 9)

  -0.1 +0.3 +0.1

-05

b.

P(x = 20) = 0.1

c.

Par > 3) = P(x = 3) + P(x =9) + P(x = 11) + P(x = 12) + Par = 20)

  =0.3+0.1+0.2+0.2+0.1

  =0.9

d.

The mean of the random variable x is:

Mean= \sum xP(x)

  \boldsymbol{=(2 \times 0.1)+(3 \times 0.3)+(9 \times 0.1)+(11 \times 0.2)+(12 \times 0.2)+(20 \times 0.1)}

=0.2+0.9+0.9+2.2+2.4+2

=8.6

e.

The standard deviation for the random variable x is:

Standard-deviation =\sqrt{\sum P(x)(x-mean)^{2}}

XP(x) 0.2 0.9 P(x) 0.1 0.3 0.1 0.2 0.2 0.9 2.2 2.4 P(x)(x-mean) 4.356 9.408 0.016 1.152 2.312 12.996 P(x)(x-mean) = 30.24 xP(

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