Let X be the average of a sample of 16 independent normal random variables with mean 0 and variance 1. Determine c such that P (X< c) = .5

To determine the value of c such that P(X < c) = 0.5, we need to find the value of c for which the cumulative distribution function (CDF) of X equals 0.5.

In this case, X is the average of a sample of 16 independent normal random variables with mean 0 and variance 1. The average of a sample follows a normal distribution with mean equal to the population mean (0 in this case) and variance equal to the population variance divided by the sample size (1/16 in this case).

The CDF of a normal distribution is given by the formula:

CDF(x) = 0.5 * (1 + erf((x - μ) / (σ * sqrt(2))))

where erf is the error function, μ is the mean, σ is the standard deviation, and x is the value for which we want to find the cumulative probability.

In this case, μ = 0 and σ = sqrt(1/16) = 1/4. Plugging these values into the CDF equation, we have:

0.5 = 0.5 * (1 + erf((c - 0) / (1/4 * sqrt(2))))

Now, we can solve for c by isolating it:

1 + erf((c - 0) / (1/4 * sqrt(2))) = 2

erf((c - 0) / (1/4 * sqrt(2))) = 1

Now, we need to find the value of c for which the error function equals 1. We can use inverse error function or the erfinv function to find this value.

c = (1/4 * sqrt(2)) * erfinv(1)

Using a scientific calculator or software that has the erfinv function, we can evaluate erfinv(1) to get:

c = (1/4 * sqrt(2)) * infinity

Since the inverse error function evaluated at 1 gives infinity, we can conclude that there's no finite value of c such that P(X < c) = 0.5.

To calculate the value of c such that P(X < c) = 0.5, we need to determine the distribution of the sample mean X.

Given that the sample consists of 16 independent normal random variables, each with a mean of 0 and variance of 1, we can use the properties of the normal distribution to find the distribution of X.

The mean of the sample mean, denoted as μX, will still be 0, as the mean of each individual random variable is 0.

However, the variance of the sample mean, denoted as σX^2, can be calculated by dividing the variance of an individual random variable by the sample size (n).

Since the variance of each random variable is 1, and the sample size is 16, the variance of the sample mean is σX^2 = 1/16 = 0.0625.

Now, we can use the properties of the standard normal distribution (mean = 0, variance = 1) to find the value of c.

The probability P(X < c) corresponds to the cumulative distribution function (CDF) of X.

To find the value of c such that P(X < c) = 0.5, we need to find the z-score value from the standard normal distribution.

Using a standard normal distribution table or calculator, we can find that the z-score value corresponding to a cumulative probability of 0.5 is 0.

Since z = (X - μX) / σX, we can rearrange the equation to find c:

c = 0 * σX + μX = 0 * √(0.0625) + 0 = 0.

Therefore, the value of c such that P(X < c) = 0.5 is c = 0.