According to a college survey, 22% of all students work full time. Find the mean and standard deviation for the number of students who work full time in samples of size 16.
According to a college survey, 22% of all students work full time. Find the mean for the number of students who work full time in samples of size 16.
3.52 is the answer to this question
Well, 22% of students work full time... Are they preparing to become circus clowns too? Anyway, let's calculate the mean and standard deviation for the number of students who work full time in samples of size 16.
Mean:
The mean is calculated by multiplying the probability of success (p) by the sample size (n). In this case, the probability of a student working full time is 0.22 and the sample size is 16.
Mean = p * n = 0.22 * 16 = 3.52
So, the mean number of students who work full time in samples of size 16 is 3.52.
Standard Deviation:
The standard deviation of a binomial distribution is calculated using the formula √(p * (1-p) * n). Using the same values, we can calculate it.
Standard Deviation = √(p * (1-p) * n) = √(0.22 * (1-0.22) * 16) ≈ √(0.17472) ≈ 0.418
Therefore, the standard deviation for the number of students who work full time in samples of size 16 is approximately 0.418. Keep those clowns in line!
To find the mean and standard deviation for the number of students who work full time in samples of size 16, we need to apply the formulas for these statistics.
Mean (μ) is calculated as:
μ = p * n
where p is the population proportion and n is the sample size.
Standard Deviation (σ) is calculated as:
σ = √(p * (1 - p) / n)
where p is the population proportion and n is the sample size.
In this case, the population proportion is given as 22% or 0.22, and the sample size is 16.
1. Calculate the mean:
μ = 0.22 * 16
μ = 3.52
Therefore, the mean number of students who work full time in samples of size 16 is 3.52.
2. Calculate the standard deviation:
σ = √(0.22 * (1 - 0.22) / 16)
σ = √(0.22 * 0.78 / 16)
σ = √(0.1716 / 16)
σ = √0.010725
σ ≈ 0.1036 (rounded to four decimal places)
Therefore, the standard deviation for the number of students who work full time in samples of size 16 is approximately 0.1036.
A 27 year old woman decides to pay $156 for a 1 year life insurance policy with coverage of $100,000. The probability of her living through the year is 0.9995. Determine her expected value for the insurance policy show work
mu = np
mean = np
= 0.22 *16 =3.52
Standard Deviation = Sqrt(n PQ)
= Sqrt(16 *.22 *.78)
= 1.656985