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

40%

of companies in Country A that have three or more female board directors?

Answer #1

Solution

Given that,

p = 0.36

1 - p = 1-0.36=0.64

n = 100

= p =0.36

= [p ( 1 - p ) / n] = [(0.36*0.64) /100 ] = 0.048

= P( 0.30<<0.40 )= P[(0.30 -0.36) / 0.048< ( - ) / < (0.40 -0.36) /0.048 ]

= P(-1.25 < z <0.83 )

= P(z <0.83 ) - P(z <-1.25 )

Using z table

=0.7967-0.1056

=0.6911

probability= 0.6911

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