The random variable X has a binomial distribution with the probability of a success being 0.2 and the number of independent trials is 15. The random variable xbar is the mean of a random sample of 100 values of X. Find P(xbar<3.15).

suppose x is a random variable best described by a uniform probability distribution with c=20 and d=60. Find P(x≤20)

To find P(xbar<3.15), we need to use the central limit theorem to approximate the distribution of xbar.

The central limit theorem states that, for a large enough sample size, the distribution of the sample mean xbar will be approximately normal, regardless of the shape of the original distribution.

In this case, since we have a binomial distribution, we can use the mean (μ) and standard deviation (σ) of the original distribution to calculate the mean (μxbar) and standard deviation (σxbar) of the sample mean xbar.

The mean of the original distribution is given by μ = np, where n is the number of trials and p is the probability of success. In this case, n = 15 and p = 0.2, so μ = 15 * 0.2 = 3.

The standard deviation of the original distribution is given by σ = sqrt(np(1-p)). In this case, σ = sqrt(15 * 0.2 * (1-0.2)) = 1.8.

The mean of the sample mean xbar is the same as the mean of the original distribution, which is μxbar = μ = 3.

The standard deviation of the sample mean xbar is given by σxbar = σ/sqrt(n), where n is the sample size. In this case, n = 100, so σxbar = 1.8/sqrt(100) = 0.18.

Now, we can standardize the value xbar = 3.15 using the approximation formula for the normal distribution:

z = (x - μxbar) / σxbar

Substituting the values, we have:

z = (3.15 - 3) / 0.18 = 0.8333

Now, we need to find the probability that a standard normal random variable is less than 0.8333.

Using a standard normal distribution table or a calculator, we can find that P(Z < 0.8333) is approximately 0.7967.

Therefore, the probability that xbar is less than 3.15 is approximately 0.7967.