$Let X_{1}, X_{2}, X_{3} and X_{4}$ be independent normal random variables with zero mean and unit variance. The probability that $X_{4}$ is the smallest among the four is .

0.25

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Solution

The correct option is A 0.25
$Since X_{1}, X_{2}, X_{3} and X_{4}$ are independent normal random variable with zero mean and unit variance then

$P (X_{4} is smallest) = P (X_{3} is smallest)$

$= P (X_{2} is smallest) = P (X_{1} is smallest) = P$

$P (X_{4} is smallest) = \frac{P}{4 P} = \frac{1}{4} = 0.25$

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Let X1, X2, X3 and X4 be independent normal random variables with zero mean and unit variance. The probability that X4 is the smallest among the four is . 0.25

The correct option is A 0.25Since X1,X2,X3 and X4 are independent normal random variable with zero mean and unit variance then P(X4 is smallest)=P(X3 is smallest) =P(X2 is smallest)=P(X1 is smallest)=P P(X4 is smallest)=P4P=14=0.25

$Let X_{1}, X_{2}, X_{3} and X_{4}$ be independent normal random variables with zero mean and unit variance. The probability that $X_{4}$ is the smallest among the four is .

0.25