For a random variable x- -x- following normal distribution- the mean is - -100- If the probability is P- for x- 110- Then the probability of x lying between 90 and 110 i-e- P90- x- 110 and equal to

Given nbsp; nbsp; nbsp;P(x gt;110)=α nbsp; nbsp; nbsp;... (1) Now P(90 lt;x lt;110)=2P(100 lt;x lt;110) nbsp; nbsp; nbsp;= ...

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Engineering Mathematics

Normal Distribution Properties

For a random ...

Question

For a random variable $x (- \infty < x < \infty)$ following normal distribution, the mean is $μ = 100.$ If the probability is $P = α for x \geq 110.$ Then the probability of x lying between 90 and 110 i.e. $P (90 \leq x \leq 110)$ and equal to

1 - 2 α

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1 - α

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1 - α / 2

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2 α

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Solution

The correct option is A $1 - 2 α$
Given $P (x > 110) = α$ ... (1)

Now

$P (90 < x < 110) = 2 P (100 < x < 110)$

$= 2 [P (100 < x < \infty) - P (x \geq 110)]$

$= 2 [\frac{1}{2} - p (x > 110)]$

$= 2 [\frac{1}{2} - α]$

$= 1 - 2 α$

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For a random variable x(−∞<x<∞) following normal distribution, the mean is μ=100. If the probability is P=α for x≥110. Then the probability of x lying between 90 and 110 i.e. P(90≤x≤110) and equal to

The correct option is A 1−2αGiven P(x>110)=α ... (1) Now P(90<x<110)=2P(100<x<110) =2[P(100<x<∞)−P(x≥110)] =2[12−p(x>110)] =2[12−α] =1−2α

For a random variable $x (- \infty < x < \infty)$ following normal distribution, the mean is $μ = 100.$ If the probability is $P = α for x \geq 110.$ Then the probability of x lying between 90 and 110 i.e. $P (90 \leq x \leq 110)$ and equal to

The correct option is A $1 - 2 α$
Given $P (x > 110) = α$ ... (1)

Now

$P (90 < x < 110) = 2 P (100 < x < 110)$

$= 2 [P (100 < x < \infty) - P (x \geq 110)]$

$= 2 [\frac{1}{2} - p (x > 110)]$

$= 2 [\frac{1}{2} - α]$

$= 1 - 2 α$