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 30% and 38% of companies in Country A that have three or more female board directors?

Answer #1

p = 0.36

n = 100

= p = 0.36

= sqrt(p(1 - p)/n)

= sqrt(0.36 * (1 - 0.36)/100)

= 0.048

a) P(0.3 < < 0.38)

= P((0.3 - )/ < ( - )/) < (0.38 - )/)

= P((0.3 - 0.36)/0.048 < Z < (0.38 - 0.36)/0.048)

= P(-1.25 < Z < 0.42)

= P(Z < 0.42) - P(Z < -1.25)

= 0.6628 - 0.1056

= 0.5572

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