Respuesta :
Answer:
(a) The probability of no women being promoted is 17%.
(b) The probability of only one women being promoted is 6%.
(c) The probability of having 2 women or more in the promotions is 32%, so it is not suspicious.
Step-by-step explanation:
This is a case of combination (no replacement and order doesn't matter).
(a) The probability that no woman is promoted is equal to the probability of a man being choose for every promotion:
[tex]P(no\, woman)=P(x_1=m)*P(x_2=m)*P(x_3=m)*P(x_4=m)*P(x_5=m)*P(x_6=m)\\\\P(no\, woman)=(75/100)*(74/99)*(73/98)*(72/97)*(71/96)*(70/95)\\\\P(no\, woman)=\frac{1.45*10^{11}}{8.58*10^{11}} =0.17[/tex]
The probability of no women being promoted is 17%.
(b) The probability of only one woman being promoted is
[tex]P(one\, woman)=P(x_1=w)*P(x_2=m)*P(x_3=m)*P(x_4=m)*P(x_5=m)*P(x_6=m)\\\\P(one\, woman)=(25/100)*(75/99)*(74/98)*(73/97)*(72/96)*(71/95)\\\\P(one\, woman)=\frac{5.18*10^{10}}{8.58*10^{11}} =0.06[/tex]
The probability of only one women being promoted is 6%.
(c) The expected number of women in the 6 promotions can be calculated as [tex]E(w)=p_w*6=(25/100)*6=1.5[/tex].
This expected value, as approximated by a binomial distribution with p=0.25 (chances of picking a woman) and n=6, is
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{6*0.25*0.75)}=\sqrt{1.125}=1.06[/tex]
If we compute the z-value for 2 woman, and approximating by the central limit theorem, we can calculate the probability of this event.
[tex]z=\frac{x-\mu}{\sigma}=\frac{2-1.5}{1.06}= 0.47[/tex]
Then P(z>-.47)=0.32 or 32%, what can be interpreted as the probability of having 2 or more women in the promotions.
We can conclude that it is not suspicious to have 2 women selected for promotions.
