# Q 1) Use Binomial Probabilities Table VII from Appendix section A-22(attached) to find probability that for the certain number of trials N and probability of success in each trial p number of...

Q 1) Use Binomial Probabilities Table VII from Appendix section A-22(attached)
to find probability that for the certain number of trials N
and probability of success in each trial p
number of successes will be:
1) exactly k
2) less than k
3) more than k

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

Q 2)

Table below 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.

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

(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).

(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 `

Sources: