SOLUTION: 67% of children in a school do not have cavities. Let x be the number of children in a random sample of 70 children selected from this school who do not have cavities. The mean and

Question 1169655: 67% of children in a school do not have cavities. Let x be the number of children in a random sample of 70 children selected from this school who do not have cavities. The mean and standard deviation of the probability distribution of x, rounded to two decimal places, are?
Just looking for clarification,
Mean is 67
The standard deviation is 8.1853
If not I'm very confused then :(
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(53765)   (Show Source): You can put this solution on YOUR website!
.

The formulation of the problem is incorrect grammatically, logically and mathematically.

As formulated, it is AMBIGOUS, which is PROHIBITED in Math problems.

/\/\/\/\/\/\/\/

After reading your post, it is UNCLEAR to me,

who do not have cavities: these 70 children or that x children ?

You may answer me :   guess it on your own.

But I don't want to guess: It is Math, and I want to KNOW / (to see) it from the text - not from the context.

Answer by math_tutor2020(3835)   (Show Source): You can put this solution on YOUR website!

67% of the children do not have cavities
p = 0.67 is the probability of selecting a child who does not cavities.

n = 70 is the sample size

mu = mean
sigma = standard deviation

For a binomial distribution it has the following formulas
mu = n*p
sigma = sqrt( n*p*(1-p) )

Let's first find the mean
mu = n*p
mu = 70*0.67
mu = 46.9
The mean is exactly 46.9

Now the standard deviation
sigma = sqrt( n*p*(1-p) )
sigma = sqrt( 70*0.67*(1-0.67) )
sigma = 3.93408184968233
sigma = 3.93
The standard deviation is approximately 3.93

Answers:
mean = 46.9
standard deviation = 3.93

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