Question 238622: Can you please help with the following binomial distributions!!!!!
1. Apprx 10.3% of americans high school students drop ou bf graduation. Choose 10 students entering school at randon, find the proability
a. no more than 2 drop out
b. at least 6 graduate
c. all 10 stay and graduate
2. The % of couples where both parties are in the labor force is 52.1% choose 5 couples at random and find the proability that none of the couples have both persons working
a. more than 3 couples have both persons in the labor force
b. fewer than 2 of the couples have both parties working
3. It has been reported that 83% of federal empleyees use email, if a sample of 200 employees is selected find the mean, varience, and standard deviation of the number who use email.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Can you please help with the following binomial distributions!!!!!
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Comment: I will be using a TI calculator to get the numerical answers.
You will need a calculator or binomial charts for these problems.
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1. Apprx 10.3% of americans high school students drop out bf graduation. Choose 10 students entering school at randon, find the proability
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These are binomial with n=10, p=0.103
a. no more than 2 drop out
P(0<= x <=2) = binomcdf(10,0.103,2) = 0.9245
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b. at least 6 graduate
P(6 <= x <= 10) = 1 - binomcdf(10,0.103,5) = 0.0001735
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c. all 10 stay and graduate
P(x = 10) = binompdf(10,0.103,10) = 0.0000000001344
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2. The % of couples where both parties are in the labor force is 52.1% choose 5 couples at random and find the proability that none of the couples have both persons working
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Binomial with n=5, p=0.521
P(x = 0) = binompdf(5,0.521,0) = 0.0252
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a. more than 3 couples have both persons in the labor force
= 1 - binomcdf(5,0.521,3) = 0.2149
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b. fewer than 2 of the couples have both parties working
P(0 <= x <= 1) = binomcdf(5,0.521,1) = 0.1624----
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3. It has been reported that 83% of federal empleyees use email, if a sample of 200 employees is selected find the mean, varience, and standard deviation of the number who use email.
mean = np = 0.83*200 = 166
variance = npq = 166*0.17 = 28.22
std = sqrt(28.22) = 5.3122
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Cheers,
Stan H.
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