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 (e) below. a. What is the probability that the sample wil have between 30% and 37% of companies in Country that have three or more female board directors ? The probability in (Round to four decimal places as needed) b. The probability is 90% that the sample percentage of Country A companies having three or more female board directors will be contained within what wymmetrical limits of the population percentage? The probability is 90% that the sample percentage will be contained above and below % (Round to one decimal place as needed.) c. The probability is 68% that the sample percentage of Country A companies having three or more female board directors wil be contained within what symmetrical limits of the population percentage? The probability is 68% that the sample percentage will be contained above and below %. (Round to one decimal place as needed.)

Answer 1

a) Given , p= 0.36 ( population proportion)

sample size , n= 100

Let be the sample proportion

р

To find , P( 0.30 < < 0.37)

р

By Central limit theorem , the sampling distribution of sample proportion follow Normal with

mean = 0.36

and standard error =

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

That is

P-N(0.36, 0.0482

then

2 = P-0.36 0.048 N(0,1)

P( 0.30 <   < 0.37)

р

0.30 – 0.36 = PG 0.048 0.37 – 0.36 0.048

= P( -1.25 < z < 0.21 )

= P( -1.25 < z < 0) +P( 0< z < 0.21)

= 0.3944 + 0.0832 ( from z table )

= 0.4776

Answer : 0.4776

Note : From z table , 0 to z

P( -1.25 < z < 0)= P( 0 < z < 1.25) = 0.3944

P( 0 < z < 0.21) =0.0832

b) The interval within 90% of sample proportion will lie

pĖ zc* (1 - p)/n

For 90% probability , zc = 1.645

That is P( - 1.645 < z < 1.645)= 0.90

Therefore we get ,

0364 164.5 $11118

=13641072

= (0.281 , 0.439)

= (28.1% , 43.9%)

The probability is 90% that the sample percentage will be contained above 28.1% and below 43.9%

c)

For 68% probability , zc = 1

That is P( - 1 < z < 1)= 0.68

Therefore we get the ,interval within 68% of sample proportion will lie

036 $ 1108

=136士008

= (0.312 , 0.408)

= (31.2% , 40.8%)

The probability is 68% that the sample percentage will be contained above 31.2% and below 40.8%