Let X be the average of a sample of size 25 independent normal random variables with mean
0 and variance 1. P[[X<c]=0.5
(a) Determine c such that
(b) Now assume that the observations may not be normally distributed, but they still are
independent with mean 0 and variance 1. Find an upper bound on the probability that P[[X<c]
with c taken from the first part of the question.
(a) To determine the value of c such that P[X < c] = 0.5, we can use the fact that X follows a Student's t-distribution with n-1 degrees of freedom when the sample size is n.
Since we have a sample of size 25, X follows a Student's t-distribution with 24 degrees of freedom. To find the value of c, we need to find the quantile for the t-distribution with the desired probability.
Using a t-table or a statistical calculator, we can find the quantile for the t-distribution with 24 degrees of freedom such that P[T < c] = 0.5. In this case, c will be the two-tailed critical value. For a two-tailed test, the area in each tail is 0.25. Therefore, we need to find the t-value corresponding to the cumulative probability of 0.25 + 0.5 = 0.75 in the t-distribution.
The critical value for a two-tailed test at this cumulative probability is approximately 0.674. Thus, c = 0.674.
(b) If we assume that the sample observations may not be normally distributed, we cannot use the t-distribution to find an exact upper bound on P[X < c]. However, we can still find a conservative upper bound using Chebyshev's inequality.
Chebyshev's inequality states that the probability that a random variable X deviates from its mean by more than k standard deviations is at most 1/k^2.
In this case, X has a mean of 0 and a variance of 1. So, the standard deviation of X is 1. To find an upper bound on P[X < c], we can substitute k as (c - 0)/1 = c.
Setting up the inequality, we have P[|X| > c] ≤ 1/c^2.
Since we want to find an upper bound on P[X < c], we can rewrite the inequality as P[X < c] ≤ 1 - P[|X| > c].
Using the inequality, we can say that P[X < c] is less than or equal to 1 - P[|X| > c], which is less than or equal to 1 - 1/c^2.
So, we have P[X < c] ≤ 1 - 1/c^2, which provides an upper bound on the probability with the value of c found in part (a).
To solve this problem, we need to use the properties of the normal distribution and the central limit theorem.
(a) Determine c such that P[X < c] = 0.5
Since each random variable in the sample follows a normal distribution with mean 0 and variance 1, the average X also follows a normal distribution with mean 0 and variance 1/N, where N is the sample size.
To find the value of c, we need to find the z-score corresponding to the desired probability. The z-score represents the number of standard deviations away from the mean.
Since the probability P[X < c] is 0.5, we can find the z-score by looking up the cumulative distribution function (CDF) of the standard normal distribution. For a standard normal distribution, the CDF gives the probability of a random variable falling below a certain value.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to P[Z < z] = 0.5 is z = 0.
Now, we can use the relationship between the z-score and c to find the value of c:
z = (c - mean) / standard deviation
Since the mean is 0 and the standard deviation is 1/sqrt(N), we have:
0 = (c - 0) / (1/sqrt(N))
0 = c / (1/sqrt(N))
Simplifying, we get:
c = 0 * (1/sqrt(N))
c = 0
Therefore, c = 0 is the value that satisfies P[X < c] = 0.5 for a sample size of 25 independent normal random variables with mean 0 and variance 1.
(b) Find an upper bound on the probability P[X < c] with c from part (a) where the observations may not be normally distributed, but are still independent with mean 0 and variance 1.
In this case, we can use the central limit theorem to approximate the distribution of X. The central limit theorem states that the distribution of the sample mean approaches a normal distribution as the sample size increases, regardless of the shape of the underlying distribution.
Since the sample size is 25, we can safely assume that the distribution of X is approximately normal. The mean and variance of X remain as before (mean = 0 and variance = 1/N).
To find an upper bound on P[X < c], we can use the z-score approach as in part (a). However, since the observations may not be normally distributed, the actual probability may not be exactly 0.5.
To find an upper bound, we can calculate the z-score using the sample size, mean, and variance as before, and then use the standard normal distribution to find the probability P[Z < z] for the calculated z-score.
Keep in mind that this will only give us an upper bound, as the actual probability may be smaller due to the non-normality of the observations.