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Question 7.6: Consider a continuous random variable X with expectation 15 ......

Consider a continuous random variable X with expectation 15 and variance 4. Determine the smallest interval [15 − c, 15 + c] which contains at least 90 % of the values of X.

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Using Tschebyschev’s inequality (7.24)

P\left(\left|X-\mu \right|\lt c \right) \geq 1-\frac{Var\left(X\right) }{c^{2}} .  (7.24)

P\left(\left|X-\mu \right|\lt c \right) \geq 0.9=1-\frac{Var\left(X\right) }{c^{2}} ,

we can determine c as follows:

1-\frac{Var\left(X\right) }{c^{2}}=0.9

c^{2}=\frac{Var\left(X\right) }{0.1}=\frac{4}{0.1}=40

c= \pm  \sqrt{40} = \pm 6.325.

Thus, the interval is [15 − 6.325; 15 + 6.325] = [8.675; 21.325].

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