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I'm having a little difficulty with an assignment that I was just assigned, and I'm not too sure if I did it right but if anyone would like to check my answer to see if it's correct or not that would be great. Maybe you could even provide the correct way or correct answer.


Here's the question:

M1 and F1 = below 40 M2 and F2 = above 40 M=Male F=Female M1: 65 M2: 40 F1: 45 F2: 30?

Based on the data provided above find: 1) Probability that randomly selected person is Female and is less than 40 years old: P( Female and Below 40). 2) Probability that randomly selected person is Male or person is above 40 years old: P (Male or Above 40) 3) Conditional Probability that if selected person is female her age is below 40, conditional probability P(Below 40 if Female). Tip: in this case you should divide by number of Female in the table, not by total number of all people.


Here's my work:

For N, p, k use numbers assigned= N= 12, K=5 p=0.4

Q 2)

Table above shows number of people who work for the company, separated by age and gender. For M1,M2,F1,F2 use numbers assigned for you= M1=35,M2=25, F1=60, F2=40

below 40 above 40 Male M1 M2 Female F1 F2 Based on data in your table find different probability: 1) that randomly selected person is Female and is less than 40 years old: P( Female and Below 40). 2) Probability that randomly selected person is Male or person is above 40 years old: P (Male or Above 40) 3) Find the probability that first selected person is male above 40 and second person is female with age below 40.

(1) Given N=12, p=.4 and k=5:

(a) From the binomial distribution table P(x=5) is 0.227. We can calculate this directly as _12C_5 (.4)^5(.6)^7 using P(x=k)= _NC_k (p)^k(q)^(N-k) where N is the total of subjects, k is the number chosen from the group, p is the probability of an individual having the sought after characteristic, and q is the complement of p.

Here _(12)C_5 (.4)^5(.6)^7=792(.01024)(.0279936)~~.227 (b) To find P(x<5) we sum P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4).

From the table add .002+.017+.064+.142+.213=.438

(c) To find P(x>5) we could add P(x=6)+ ... +P(x=12). Alternatively, we can realize that P(x>5) is the complement of P(x<=5)=P(x=5)+P(x<5)=.227+.438=.665 So P(x>5)=1-.665=.335

Check: from the table add .177+.101+.042+.012+.002=.334. The table does not have entries for k=11 or 12 which accounts for the discrepancy. My calculator gives .3347914423

(2) There are 160 total employees: 60 male and 100 female. 95 employees are less than 40, while 65 employees are older than 40.

(a) Find P(F and <40) -- there are 60 females that are younger than 40, so the probability is 60/160=3/8 .

Alternatively, the probability of being younger than 40 and female is (95/160)*(60/95)=60/160=3/8 (95/160 is the probability of being younger than 40 while 60/95 is the probability of being female given that you are under 40.

(b) Find P(M or >40). These events are not disjoint (or not mutually exclusive) so we use P(A "or"B)=P(A)+P(B)-P(Ann B) Here P(M)=60/160 and P(>40)=65/160 while the intersection is 25/160 so:

P(M or >40)=60/160+65/160-25/160=5/8 Check: There are 60 males and 65 employees over 40; 25 of those employees were already counted so we have 60+40=100 employees that are male or over 40 and the probability is 100/160=5/8 (c) Selecting two people, find P(f,<40|m,>40):

These are not independent events, assuming that the male is removed from the list after selection.

P(M>40)=25/160=5/32. Then P(F<40)=60/159 (there are 60 females under 40 to choose, but only 159 people to choose from after removing the male.)

5/32*60/159=25/424

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  • $\begingroup$ Your question in bold text made no sense to me so I stopped reading at that point. Could you make its meaning clearer? $\endgroup$ – Paul Aug 14 '15 at 15:31
  • $\begingroup$ The work done has no clear relationship to the questions asked at the beginning. In particular, there is no question asked whose answer involves the binomial distribution. $\endgroup$ – André Nicolas Aug 14 '15 at 15:45
  • $\begingroup$ Can you present the data in tabular form? I'm guessing this is a chi-squared test for a 2 x 2 table (four cells). If so, there should be an example or formula for that in your textbook. If that is so, and you still don't understand, please tell us what the text has to say about 2 x 2 tables. Notation and rationales differ widely among texts, so some background would be needed in order to be really helpful at that point. Alternatively, this could be framed as a test of equality of two binomial proportions, and we'd need to know that. $\endgroup$ – BruceET Aug 15 '15 at 22:04

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