A random variable X takes the values – 2, -1, 1 and 2 with probabilities 1−a4,1+2a4,1−2a4 and 1+a4 respectively then

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a can have any real value

−12≤a≤12

−1≤a≤1

14≤a≤13

answer is B.

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Detailed Solution

0≤1−a4≤1⇒0≤1−a≤4⇒0−1≤−a≤3⇒-3≤a≤10≤1+2a4≤1⇒0≤1+2a≤4⇒−1≤2a≤3⇒−12≤a≤32 0≤1−2a4≤1⇒0≤1−2a≤4⇒−1≤−2a≤3⇒-32≤a≤120≤1+a4≤10≤1+a≤40≤a≤3therefore a∈−12,12

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