Find the Z score that correspond to 61.4% of the distribution area to the right

Question

Question

math Statistics-And-Probability

Answer 1

Answer #1

solution

Using standard normal table,

P(Z > z) = 61.4%

= 1 - P(Z < z) = 0.614

= P(Z < z) = 1 - 0.614

= P(Z < z ) = 0.386

= P(Z < -0.29 ) = 0.0.386

z = -0.29 ( using z table )

Answer 2

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