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A random variable is normally distributed with a mean of μ = 50 and a standard...

A random variable is normally distributed with a mean of μ = 50 and a standard deviation of σ = 5. What is the probability that the random variable will assume a value that is less than 40? Make sure your answer is between 0 and 1, round to four digits.

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we have mean = 50 and standard deviation = 5

we have to find the probability that a value is less than 40

Using the formula P(X<x) = P(z<(ar{x}-mu)/sigma)

setting ar{x} = 40, mu = 50 , sigma=5

we get

P(X <40) P (40-50)/5)- Pz < (-10/5))

it gives us, P(z < (-10/5))-P(z <-2)

Using the identity P(z<-a) = 1-P(z<a)

we can write it as P(z <-2) 1-P(z < 2) = 1-0.9772 = 0.0228 (using z distribution table)

So, the required probability that a value will be less than 40 is 0.0228 (rounded to 4 decimal places)

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