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Assume that xis a random variable best described by a uniform distribution with c=10andd=90.

a. Findf(x).

b. Find the mean and standard deviation of x.

c. Graph the probability distribution for xand locate its mean and theintervalon the graph.

d. FindP(x≤60).

e. FindP(x≥90).

f. FindP(x≤80).

g. FindP(μ-σ≤x≤μ+σ).

h. FindP(x>75).

Short Answer

Expert verified

a.fx=180;10≤x≤90

b. The mean isμ=50 and the standard deviation is σ=23.094.

c. The graph is a symmetrical curve.

d. P(x≤60)=0.625

e. P(x≥90)=0

f. P(x≤80)=0.875

g. P(μ-σ<x<μ+σ)=0.57735

h. P(x>75)=0.57735

Step by step solution

01

Given information

X is auniformrandom variable.

c=10d=90

02

Calculate f(x)

a.

The p.d.f of X is

fx=1d-c;c≤x≤d=190-10=180;10≤x≤90

Hence, f(x)=180;10≤x≤90.

03

Calculate mean and standard deviation

b.

Mean=μ

μ=c+d2=10+902=1002=50

Standard deviation=σ

σ=d-c12=90-1012=8012=23.094

Hence, the mean isμ=50and standard deviation is σ=23.094.

04

Plot the graph

c.

μ±2σ=50±2×23.09450±46.188=3.812,96.188

05

Calculate P(x≤60)

d.

Px≤60=∫1060fxdx=∫1060180dx=180∫1060dx=180×50=58=0.625

Hence, P(x≤60)=0.625.

06

Calculate P(x≥90)

e.

Px≥90=1-Px<90=1-∫1090fxdx=1-∫1090180dx=1-180×80=1-1=0

Hence, P(x≥90)=0.

07

Calculate P(x≤80)

f.

Px≤80=1-∫8090fxdx=1-∫8090180dx=1-180×10=1-18=0.875

Hence, P(x≤80)=0.875.

08

Calculate P(μ-σ≤x≤μ+σ)

g.

Pμ-σ<x<μ+σ=∫μ-σμ+σfxdx=∫26.90673.094180dx=180×46.188=0.57735

Hence,role="math" localid="1660281305747" P(μ-σ<x<μ+σ)=0.57735.

09

Calculate

h.

Px>75=∫7590fxdx=∫7590180dx=180×15=316=0.57735

Hence, P(x>75)=0.57735.

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