The Center of Statistics for Health reported that 30% of adult Americans smoke. Consider a random sample of 150 adults selected randomly in this population. What is the probability that the number of non-smokers in this sample is:
a) 0
b)60 or more
c)Between 50 and 90
d)40 or less
e)Exactly 50
a) 0
b)60 or more
c)Between 50 and 90
d)40 or less
e)Exactly 50
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The exact situation is Binomial distribution with p = 0.7, q = 0.3, n = 150
I have put p = 0.7 because the question is about non-smokers.
Only the first part should be done using the exact situation.
a) P(x = 0) = 0.7^150
This is so low that it is almost zero. In fact I wouldn't mark down anyone who said it was effectively zero.
For the remaining parts use the Normal distribution approximation to the Binomial. Use the same mean = n*p =150*0.7 = 105 and standard deviation = sqrt(n*p*q) = 5.61
Remember to change to the Standard Normal variable using the formula
z = (x - mean) / s.d.
Then look up z values in tables.
Don't forget the continuity correction so for part (e) you will look for the probability of between x = 49.5 and 50.5
I have put p = 0.7 because the question is about non-smokers.
Only the first part should be done using the exact situation.
a) P(x = 0) = 0.7^150
This is so low that it is almost zero. In fact I wouldn't mark down anyone who said it was effectively zero.
For the remaining parts use the Normal distribution approximation to the Binomial. Use the same mean = n*p =150*0.7 = 105 and standard deviation = sqrt(n*p*q) = 5.61
Remember to change to the Standard Normal variable using the formula
z = (x - mean) / s.d.
Then look up z values in tables.
Don't forget the continuity correction so for part (e) you will look for the probability of between x = 49.5 and 50.5