Question

a normal shaped distribution has a mean =200 and a standard deviation =30. What are the z score values that form the boundaries for the middle 95% of the distribution?

Answer 1

Solution :-

Given :- for the Middle 95 % of the distribution is,

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

P ( -z < Z < z ) = 0.95

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

Answer = The z score values that form the boundaries for the  Middle 95 % of the distribution are +/- 1.96.

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Given - Mean = 200, standard deviation = 30

Boundaries to be formed are

for z = 1.96

By using z score formula,

X1 = z * σ + u

X1 = 1.96 * 30 + 200

X1 = 258.8

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for z = - 1.96

By using z score formula,

X2 = z * σ + u

X2 = - 1.96 * 30 + 200

X2 = 141.2