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प्रश्न
For the following probability density function (p. d. f) of X, find P(X < 1) and P(|x| < 1)
`f(x) = x^2/18, -3 < x < 3`
= 0, otherwise
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उत्तर
We have
P(X < 1) = `int_-3^1 f(x) dx`
= `int_-3^1 x^2/18dx`
= `[x^3/54]_-3^1`
= `1/54 - ((-27)/54)`
= `28/54`
= `14/27`
= 0.5185
Now |X| < 1
`\implies` ± X < 1
∴ X < 1, – X < 1,
i.e. X > – 1
i.e. – 1 < X < 1
∴ Required Probability = `int_-1^1 f(x)dx`
= `int_-1^1 x^2/18dx`
= `[x^3/54]_-1^1`
= `1/54 + 1/54`
= `2/54`
= `1/27`
= 0.03704
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