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P(|X|< 1) = ?

A1/2

B1/8

C1/4

D3/4

Answer:

C. 1/4

Read Explanation:

02f(x)dx=1\int_0^2 f(x)dx=1

02kx=1\int_0^2 kx =1

k[x22]02=1k[\frac{x^2}{2}]_0^2=1

k[42]=1k[\frac{4}{2}]=1

2k=12k=1

k=12k=\frac{1}{2}

P(|X|<1) = P(0<x<1)

=01f(x)dx=\int_0^1 f(x)dx

=k01xdx=k\int_0^1xdx

=k×12=k \times \frac{1}{2}

=12×12=14=\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}


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മധ്യാങ്കം കാണുക

mark

50-59

60-69

70-79

80-89

Frequency

10

8

30

2