Question 1091257: you are told that there is an 80% chance that a prospective employer will check the educational background of a job applicant. 75 job applications are randomly selected.
A. find the probability that at least 50 of the applicants have their educational backgrounds checked
B. find the probability that exactly 50 of the applicants have their educational backgrounds checked
Found 2 solutions by Boreal, mathmate:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! at least 50 would be 50 upward
the mean is np=75*0.80=60
the variance is np(1-p)=60*0.2=12
the sd is sqrt (12)=3.464
use continuity correction factor of 49.5
(49.5-60)/3.464=-3.03
This is the probability z >=-3.03, which is 0.9976
Exactly 50 is 75C50(0.8)^50*0.2^25=0.0025
Answer by mathmate(429)  (Show Source):
You can put this solution on YOUR website! Question:
you are told that there is an 80% chance that a prospective employer will check the educational background of a job applicant. 75 job applications are randomly selected.
A. find the probability that at least 50 of the applicants have their educational backgrounds checked
B. find the probability that exactly 50 of the applicants have their educational backgrounds checked
Solution:
Since the probability of success (checking educational background) is constant throughout, sampling is random, and size of sample is known, we can model using the binomial distribution.
P(x)=C(n,x)p^x*(1-p)^(n-x)
with C(n,x)=n!/(x!(n-x)!) number of combinations of x objects taken out of n.
Here, n=75, x=50, p=0.8
(B) exactly 50 applicants (out of 75) have their backgrounds checked:
P(X=50)=C(75,50)*(0.8^50)(0.2^25)
=52588547141148893628*1.427247692705967*10^-5*3.355443200000008*10^-18
= 0.00251849
(A) At least 50 applicants
P(X>=50)=P(X=50)+P(x=51)+P(X=52)+...+P(x=75)
=0.0025185+0.0049382+0.0091167+0.0158252+0.0257893
+0.0393872+0.0562675+0.0750233+0.0931324+0.1073390
+0.1144950+0.1126180+0.1017195+0.0839589+0.0629692
+0.0426253+0.0258335+0.0138807+0.0065321+0.0026507
+0.0009088+0.0002560+0.0000569+0.0000094+0.0000010
+0.0000001
=0.9978525
Since the above calculation is quite tedius, in most cases, solution is obtained using the normal approximation.
mean=75*0.8=60
variance=75*0.8*0.2=12
standard deviation = sqrt(12)
Applying the continuity correction,
Z=(49.5-60)/sqrt(12)=-3.0311
P(Z>49.5)
=1-P(Z>49.5)
=1-0.001218367
=0.9987816
with an error of -0.093%
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