What is the value of z if the area between -z and z is 0.802 c....

Question

Question

math Statistics-And-Probability

Answer 1

Answer #1

Solution:-

Given that

Using standard normal table,

P( -z < Z < z) = 0.80

= P(Z < z) - P(Z <-z ) = 0.0.80

= 2P(Z < z) - 1 = 0.0.80

= 2P(Z < z) = 1 + 0. 0.80

= P(Z < z) = 1.80 / 2

= P(Z < z) = 0.90

= P(Z < 1.28 ) = 0.90

= z ± 1.28

Answer 2

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