SOLUTION: So ... I am stuck ... again ... I am looking forward to the end of this paper! Here we go ... use the normal approximation to the binomial method, to find the probability that 20

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Question 880572: So ... I am stuck ... again ... I am looking forward to the end of this paper!
Here we go ... use the normal approximation to the binomial method, to find the probability that 20 or fewer of 100 people randomly selected use the vaccination programme. (We know that 25% of the staff make use of the free vaccinations)
The sample size is equal to? (I have 100)
The probability of a person using the vaccination programme is? (I have 0.02)
The mean value (�X) is equal to?
The standard deviation (σX) is equal to?
the z-value is equal to?
The probability of selecting 20 or fewer people who use the vaccine programme is equal to (to 4 decimal places)
Thanks so much for you help :-)
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
This is a sampling distribution of proportion, we know
sample size is 100 and P(probability of taking vaccination) = .25 and Q(probability of not taking vaccination) = .75
mean of p is .25 and standard deviation of p is square root (PQ/ n) where n is the sample size, therefore
standard deviation of p is square root ((.25 * .75)/ 100) = 0.04330127
now 20/100 = .20
calculate the z value = (.20 - .25) / 0.04330127 = −1.154700543
consult table of z values for probability associated with the z value
therefore P(X<20) is 0.1251