The following join probability data apply to fatigue test to run on bronze strips. X represent to failure  (in 105) when alternate strips are bent at a high level of deflection. Y represent the same at a lower deflection level.

X/Y 20 30 40 50
4 0.01 0.03 0.05 0.02
5 0.03 0.1 0.08 0.04
6 0.02 0.08 0.12 0.11
7 0.02 0.04 0.07 0.18
  1. Find the marginal probability distribution for X and Y
  2. Determine the conditional probability distribution of Y gives X = 5
  3. Are x and Y independent

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Given

X/Y 20 30 40 50 P(X=x)
4 0.01 0.03 0.05 0.02 0.11
5 0.03 0.1 0.08 0.04 0.25
6 0.02 0.08 0.12 0.11 0.33
7 0.02 0.04 0.07 0.18 0.31
P(Y=y) 0.08 0.25 0.32 0.35 1

Now,

Marginal probability of X = p(X = xi) = \(\sum_{i=4}^n p(x_i y_i)\)

∴P(X = 4) = P(X=4, Y=20) + P(X = 4, Y = 30) + P(X = 4, Y = 40) + P(X = 4, Y = 50)

= 0.11

∴P(X = 5) = P(X=5, Y=20) + P(X = 5, Y = 30) + P(X = 5, Y = 40) + P(X = 5, Y = 50)

= 0.25

∴P(X = 6) = P(X=6, Y=20) + P(X = 6, Y = 30) + P(X = 6, Y = 40) + P(X = 6, Y = 50)

= 0.33

∴P(X = 7) = P(X=7, Y=20) + P(X = 7, Y = 30) + P(X = 7, Y = 40) + P(X = 7, Y = 50)

= 0.31

Marginal probability of Y = p(Y = yi) = \(\sum_{i=4}^n p(x_i y_i)\)

∴P(Y = 20) = P(X=4, Y=20) + P(X = 5, Y = 20) + P(X = 6, Y = 20) + P(X = 7, Y = 20)

= 0.08

Similarly,

P(Y = 30) = 0.25

P(Y = 40) = 0.32

P(Y = 50) = 0.35

Answer b):

The conditional probablity function of Y given the value of X = 5 is designed as follow:

P(X=5 / Y = 20) = \(\frac{P(X=5 \enspace / \enspace Y = 20)}{P(Y = 20)}\)

=  \(\frac{0.03}{0.08} = \frac{3}{8}\)

P(X=5 / Y = 30) = \(\frac{P(X=5 \enspace / \enspace Y = 30)}{P(Y = 30)}\)

=  \(\frac{0.1}{0.25} = \frac{2}{5}\)

P(X=5 / Y = 40) = \(\frac{P(X=5 \enspace / \enspace Y = 40)}{P(Y = 40)}\)

=  \(\frac{0.08}{0.32} = \frac{1}{4}\)

P(X=5 / Y = 50) = \(\frac{P(X=5 \enspace / \enspace Y = 50)}{P(Y = 50)}\)

=  \(\frac{0.04}{0.35} = \frac{4}{35}\)\

Answer c):

Y is independent variable.

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