Question

In: Statistics and Probability

Consider a distribution with the density function f(x) = x^2/3 for −1 ≤ x ≤ 2....

Consider a distribution with the density function f(x) = x^2/3 for −1 ≤ x ≤ 2.
(a) Randomly pick a sample of size 20 from this distribution, find the probability that there are 2 to 4 (inclusive) of these taking negative values.
(b) Randomly pick an observation X from this distribution, find the probability that it is between 1.2 and 1.4, i.e., P (1.2 < X < 1.4).
(c) Randomly pick a sample of size 40 from this distribution, and let X(bar) be the sample mean, find the probability that it is between 1.2 and 1.4, i.e., find P(1.2 < X(bar) < 1.4). How does this complare to your answer in part (b)? Can you explain?
(d) Randomly pick a sample of size 4 from this distribution, find the probability that the largest one is larger than 1.8.

Expert Solution

Let the pdf of X be

(a) Randomly pick a sample of size 20 from this distribution, find the probability that there are 2 to 4 (inclusive) of these taking negative values.

The probability that a randomly picked observation from this distribution is negative is

Let Y be the number of values from a random sample of 20 taken from this distribution, taking negative values. We can say that Y has a Binomial distribution, with parameters, number of trials (size of the randomly selected sample) n=20 and the success probability (The probability that a randomly picked observation from this distribution is negative) p=1/9

The probability that Y=y of these taking negative values is given by

the probability that there are 2 to 4 (inclusive) of these taking negative values is

ans: the probability that there are 2 to 4 (inclusive) of these taking negative values is 0.6048

(b) Randomly pick an observation X from this distribution, find the probability that it is between 1.2 and 1.4, i.e., P (1.2 < X < 1.4).

the probability that a randomly picked observation from this distribution between 1.2 and 1.4 is

ans: the probability that a randomly picked observation from this distribution between 1.2 and 1.4 is 0.1129

(c) Randomly pick a sample of size 40 from this distribution, and let X(bar) be the sample mean, find the probability that it is between 1.2 and 1.4, i.e., find P(1.2 < X(bar) < 1.4). How does this compare to your answer in part (b)? Can you explain?

The expected value of X is

The expected value of is

The standard deviation of X is

Randomly pick a sample of size 40 from this distribution, and let be the sample mean. Since the sample size n=40 is greater than 30, using the CLT we can say that

has a normal distribution with mean and standard deviation

the probability that the sample mean is between 1.2 and 1.4 is

ans: the probability that the sample mean is between 1.2 and 1.4 is 0.5384

This probability is much higher than the answer in part b). The value in part b) is the probability of any single observation from the distribution is between 1.2 and 1.4, where as the probability in part c) is the sample average of any given sample of size 40 from the distribution is between 1.2 and 1.4. The sample average is less variable (has lower variance) than an individual value from the distribution and hence the probability of a sample average being between 2 value is higher than the probability of a single value being between the same 2 values.

(d) Randomly pick a sample of size 4 from this distribution, find the probability that the largest one is larger than 1.8.

Let be the 4 values picked from the distribution. All the 4 have the same pdf given by

The CDF of X is

the probability that the largest one is larger than 1.8 is

ans: the probability that the largest one is larger than 1.8 is 0.6679

orchestra answered 1 year ago

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