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What are the mean and the standard deviation of a proportion?
The mean of a proportion, p, is X/n; where X is the number of instances & n is the sample size; and its standard deviation is sqrt[p(1-p)]
Find the standard deviation of the binomial distribution for which n 1000 and p 0.94?
The standard deviation is sqrt[n*p*(1-p)] = sqrt(1000*0.94*0.06) = 7.51 approx.
How do you find the sample size if you are given the confidence interval and the margin of error as well as the standard deviation?
You can't. You need an estimate of p (p-hat) q-hat = 1 - p-hat variance = square of std dev sample size n= p-hat * q-hat/variance yes you can- it would be the confidence interval X standard deviation / margin of error then square the whole thing
If X is a normal random variable with standard deviation 4.00 and if the probability that X is more than 5.52 is 1271 then what is the mean of X?
Let M be the mean. P( X > 5.52) =0.1271 Standardize X by subtracting the mean and dividing by the standard deviation. z = (52.2-M) /4 From the normal distribution tables, P( z > 1.14 )=0.1271 Therefore, (52.2-M) / 4 = 1.14 Solve for M, the mean. 52.2-M = 4(1.14) = 4.56 M=52.2-4.56 =47.64
Can normal distribution be applied on discrete data?
Yes. The normal distribution is used to approximate a binomial distribution when the sample size (n) times the probability of success (p), and the probability of failure (q) are both greater than or equal to 5. The mean of the normal approximation is n*p and the standard deviation is the square root of n*p*q.
