I can't solve this statistics problem. All I know is that it's a binomial problem.
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I can't solve this statistics problem. All I know is that it's a binomial problem.

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
This is so low that it is almost zero. In fact I wouldnt 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.z = (x - mean) / s.......
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

-
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
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