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Assume that this information comes from a standard normal distribution where the mean...

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s4leggs | Student, Undergraduate | eNoter

Posted November 2, 2010 at 11:39 PM via web

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Assume that this information comes from a standard normal distribution where the mean is 0 and the standard deviation is 1.

Find the probability between 1.05 and 2.05.

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sociality | High School Teacher | Valedictorian

Posted November 3, 2010 at 2:01 AM (Answer #1)

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Here we are given that the mean is equal to 0 and the standard deviation is 1.

Now to derive the required probability we need to use a normal distribution table. First we'll need to convert the given variables to respective z values. z = (variable - mean) / standard deviation.

So for 1.05, z = (1.05 - 0) / 1 = 1.05 and for 2.05, z = (2.05 - 0) / 1 = 2.05.

Using a normal distribution table we have the cumulative probability as 0.3531 for z = 1.05 and for 2.05 the cumulative frequency is 0.4798.

Therefore the probability of values of z lying between 1.05 and 2.05 or the variable lying between 1.05 and 2.05 is 0.4798 - 0.3531 = 0.1267.

Therefore the required probability is 0.1267.

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neela | High School Teacher | Valedictorian

Posted November 3, 2010 at 4:42 AM (Answer #2)

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A standard normal variate  has the distribution function y  = {1/sqrt(2pi)} e^(-z^2 /2).

The probability that z takes the value between 1.05 and 2.05 is given by:

P(1.05 < =z < =2.05) = Int {1/sqrt(2pi)} e^(-x^2 /2) dx .

The right side values are obtained from any standard normal tables.

P(1.05) < =z < =2.05) = P(z < 2.05) - P(1.05)

P(1.05< = z < =2.05 = 0.97982 - 0.85314 = 0.12668.

Therefore the probability that the standard normal variate is the value between 1.05 and 2.05  ie 0.12668.

 

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