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Consider the discrete probability distribution shown here.

x

10

12

18

20

p

.2

.3

.1

.4

a. Calculateμ,σ2 andσ .

b. What isP(x<15) ?

c. Calculate μ±2σ .

d. What is the probability that xis in the interval μ±2σ ?

Short Answer

Expert verified

(a) μ=15.4,σ2=18.44andσ=4.294.

(b) Px<15=0.5

(c) μ±2σ=(6.812,23.988)

(d) The required probability is 0.5.

Step by step solution

01

Given information

X is a random variable.

02

Identifying the type of random variable

a.

Mean=μ

μ=∑i=14xjPxi=10×0.2+12×0.3+18×0.1+20×0.4=2+3.6+1.8+8=15.4

Variance=σ2

σ2=∑i=14x-μ2Px=10×15.42×0.2+12-15.42×0.3+18-15.42×0.1+20-15.42×0.4=29.16×0.2+11.56×0.3+6.76×0.1+21.16×0.4=5.832+3.468+0.676+8.464=18.44

Standard Deviation=σ

σ=18.44=4.294

Hence, μ=15.4,σ2=18.44 and σ=4.294 .

03

When  p(x<15)

b.

px<15=px=10+px=12=0.2+0.3=0.5

Hence, Px<15=0.5

04

Calculating the value when  μ±2σ

c.

The interval is

μ±2σ=15.4±2×4.294=15.4±8.588=6.812,23.988

05

Calculating P (x) when μ±2σ

d.

P6.812≤x≤23.988=P0≤x≤17.176=1-Px≥17.176=1-Px=18+Px=20=1-0.1+0.4=1-0.5=0.5

Hence, the required probability is 0.5.

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91.28 92.83 89.35 91.90 82.85 94.83 89.83 89.00 84.62

86.96 88.32 91.17 83.86 89.74 92.24 92.59 84.21 89.36

90.96 92.85 89.39 89.82 89.91 92.16 88.67 89.35 86.51

89.04 91.82 93.02 88.32 88.76 89.26 90.36 87.16 91.74

86.12 92.10 83.33 87.61 88.20 92.78 86.35 93.84 91.20

93.44 86.77 83.77 93.19 81.79

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50

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Standard deviation(n-1)

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