Question

Determine the probability that a random variable with a normal distribution N(100, 10) is in the range 70 to 105.

Answer 1

Given: = 100, ² = 10, = SQRT(10) = 3.1623

To find the probability, we need to find the Z scores first.

Z = (X - )/ [/Sqrt(n)]. Since n = 1, Z = (X - )/

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P(70 < X < 105) = P(X < 105) - P(X < 70)

For P(X < 105); Z = (105 – 100)/3.1623 = 0.02

The probability for P(X < 105) = 0.5080

For P(X < 70); Z = (70 – 100)/3.1623 = -0.13

The probability for P(X < 70) = 0.4483

Therefore the required probability = 0.5080 - 0.4483 = 0.0597

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