Question

A study reports that 36% of companies in Country A have three or more female board directors. Suppose you select a random sample of 100 respondents. Complete parts (a) through (c) below.

a. What is the probability that the sample will have between 35% and 38% of companies in Country A that have three or more female board directors?

Answer #1

Using normal approximation,

P( < p ) = P(Z < ( - p) / sqrt [ p( 1 - p) / n ] )

So,

P(0.35 < < 0.38) = P( < 0.38) - P( < 0.35)

= P(Z < ( 0.38 - 0.36) / sqrt ( 0.36 ( 1 - 0.36) / 100 ) ) - P(Z < ( 0.35 - 0.36) / sqrt ( 0.36 ( 1 - 0.36) / 100 ) )

= P(Z < 0.42 ) - P(Z < -0.21)

= 0.6628 - 0.4168

= **0.2460**

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