Suppose X is a Normal random variable with with expected value 2 and standard de

Question

Suppose X is a Normal random variable with with expected value 2 and standard deviation .3. We take a random sample of size n from the distribution of X. Let _ X be the sample mean. Use R to determine the following: a) Find the probability P(X>2.2) b) Find the probability P( _ X )>2.2 when n = 4 c) Find the probability P( _ X )>2.2 when n = 16 d) What is the probability P(1.9 < _ X <2.1) when n = 16? e) What is the standard deviation of _ X when n = 16 f) What is the probability that X1 + X2 + ... +X16 < 33

Explanation / Answer

X IS GIVEN TO BE RANDOM VARIABLE

WHERE MEAN = 2

AND THE STANDARD DEVIATION = 0.3

FOR MULA TO BE USED = (X-MEAN)/STANDARD DEVIATION

A) P(X>2.2) =

For x = 2.2, z = (2.2 - 2) / 0.3 = 0.66

Hence P(x > 2.2) = P(z >0.66) = [total area] - [area to the left of 0.66]

1 - [area to the left of 0.66]

now from the z table we will take the value of z score = 0.66

    = 1 - 0.7453 = 0.2547

B)P(X>2.2) =

For x = 2.2, z = (2.2 - 2) / (0.3/SQRT(4)) = 1.33

Hence P(x > 2.2) = P(z >1.33) = [total area] - [area to the left of 1.33]

1 - [area to the left of 1.33]

now from the z table we will take the value of z score = 1.33

    = 1 - 0.9082 = 0.0918

C) N = 16

P(X>2.2) =

For x = 2.2, z = (2.2 - 2) /( 0.3/SQRT(16)) = 2.66

Hence P(x > 2.2) = P(z >2.66) = [total area] - [area to the left of 2.66]

1 - [area to the left of 2.66]

now from the z table we will take the value of z score = 2.66

    = 1 - 0.9960 = 0.004

D) P(1.9<X<2.1) =

For x = 1.9 , z = (1.9 - 2.2) /(0.3/SQRT(16) = -4 and for x = -2.1 , z = (2.1 - 2.2) / (0.3/SQRT(16) ) = - 1.33

Hence P(30 < x < 35) = P(-2.1 < z < -1.33) = [area to the left of z = -1.33] - [area to the left of -2.1]

= 0.0917 - 0.0178 = 0.0739

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