Question

Can anyone help?

Assume that the lengths of the solutions turned in for the written portion of homework 1 have mean 0.5 pages and standard deviation 0.1 pages.

a) If you pick 35 students at random in this class, what is the distribution of the 35 solutions’ average length? Assume that the lengths of different solutions are independent. (Define
shape and parameters).

b) Find the probability that the 35 solutions’ average length is at at least 0.55 pages.

Answer 1

a) The distribution of the 35 solutions' average length is a normal distribution, since we have a large enough sample size (n=35) and the lengths are assumed to be independent and identically distributed. The mean of this distribution is also 0.5 pages and the standard deviation is given by:

standard deviation = (standard deviation of individual solutions) / sqrt(n) = 0.1 / sqrt(35) ≈ 0.017

Therefore, the distribution of the 35 solutions' average length is a normal distribution with mean 0.5 and standard deviation 0.017.

b) To find the probability that the 35 solutions' average length is at least 0.55 pages, we need to calculate the z-score of this value and then find the area under the standard normal distribution curve to the right of this z-score. The z-score is given by:

z = (0.55 - 0.5) / 0.017 ≈ 2.94

Using a standard normal distribution table or calculator, we can find that the area to the right of z = 2.94 is approximately 0.0022. Therefore, the probability that the 35 solutions' average length is at least 0.55 pages is approximately 0.0022, or about 0.22%.