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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.
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.
Suppose a normal distribution has a mean of 50 and a standard deviation of 3. What is P(x≤53)?
0.84
In a random sample of 200 persons of a town 120 are found to be tea drinkers In a random sample of 500 persons of another town 240 are found to be tea drinkers Is the proportion of tea drinkers in?
Note that if the first one is multiplied by 2, you get 200*2=400 and 120*2=240. The 240 is the same as in the second test, but the population size is smaller, so the second is less. To test the significance of this difference, you would take 500-400 and divide by some standard deviation. One way to approximate this standard deviation is to say that it is from a binomial distribution with p = the average of 120/200 and 240/500. The standard deviation is sqrt[(average number of people)p(1-p)]. Use this and calculate how many standard deviations there are between the two. Look up this value on the normal distribution table to determine what will be the probability of being differernt.
Can you please tell me the variance and the standard deviation for n equals 80 and p equals 0.3?
For a binomial probability distribution, the variance is n*p*q which is 80*.3*.7 = 16.8. The standard deviation is square root of the variance which is 4.099; rounded is 4.1. The mean for a binomial probability distribution is n*p or 80*.3 or 24.
What is the standard deviation of a binomial distribution for which n 736 p 0.7?
n = 736, p = 0.7 So variance = 736*0.7*(1-0.7) = 154.56 and therefore standard deviations = sqrt(154.56) = 12.43
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
What is the standard deviation of a geometric random variable x for an event with a probability of success of 0.75?
It is sqrt[(1-p)/p2] = 0.666... recurring.
A binomial distribution has a mean of 12 and a standard deviation of 2.683 find n and p?
There is not enough information to find n & p. The mean is n*p and the std dev = sqrt (n*p*q). You have to be given n, p or q to have 2 equations 2 unknowns to solve.
List of abbreviation for 'Cpk'?
Cpk = Cp (Process Capability) + p (katayori) Japanese for deviation. Cpk = Deviation of process capability
When did P. K. Sarangapani die?
P. K. Sarangapani died on 2011-02-02.
A mean of 10.5 ounces and a standard deviation of .3 ounce. Find the proportion that have weights above 10.95 ounces?
μ = 10.5 σ = 0.3 P(x > 10.95) = P( z > (10.95-10.5) / 0.3) = P(z > 1.5) = 0.0668 or 668/1000 (using Normal probability table)
