Question

# Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1)...

Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1) what is the probability that

1. Z is between -1.23 and 1.64
2. Z Is less than -1.27 or greater than 1.74
3. For normal data with values symmetrically distributed around the mean find the z values that contain 95% of the data
4. Find the value of z such that area to the right is 2.5% of the total area under the normal curve

Solution,

Using standard normal table,

a) P( -1.23 < Z < 1.64)

= P( Z < 1.64) - P( Z < -1.23)

= 0.9495 - 0.1093

= 0.8402

b) P( -1.27 > z or z > 1.74)

= P( z < -1.27) + P( z > 1.74)

= 0.1020 + { 1 - P( z < 1.74)}

= 0.1020 + { 1 - 0.9591}

= 0.1020 + 0.0409

= 0.1429

c) Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

z = -1.96, 1.96

d) P( z > z ) = 2.5%

= 1 - P(Z < z) = 0.025

= P(Z < z) = 1 - 0.025

= P(Z < z ) = 0.975

= P(Z < 1.96 ) = 0.975

z = 1.96

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