Stats problem

If the probability of any given piece being brown is 55%, then if you took samples of 62 pieces you would expect the number of brown ones to be 34, but with normal variation, you would expect the standard deviation around the mean to be sqrt(55% x 45% x 62) = 3.9. Your observed result of 41 was 7 away from the expected mean or less than two standard deviations so that is reasonable.

Can you explain this more "Your observed result of 41 was 7 away from the expected mean or less than two standard deviations so that is reasonable. " ?

Your observed result was 41 out of 62. That is what you told us in the problem statement. You would have expected to get 55% of 62. That is 34. The difference between the actual (41) and the expected (34) is 7.

So think of tossing a coin. If you toss a coin 100 times, you would expect to get 50 heads, but not really because you know some variation around that is normal, so how much is normal? To answer this you need to know two things. The first is that if you flip 100 coins over the over again and count the number of heads, the results will form a normal distribution with a mean of 50 and a standard deviation of sqrt( P (1-P) N ) where P = 50% and N = 100. The second thing you need to know is that a basic 95% confidence interval for a normal distribution is 1.96 standard deviations. If you are outside 1.96 standard deviations, there is less than a five percent chance that the probability you used is the true probability. Statisticians like 95% and consider that a reasonable cutoff point.

In your problem, the standard deviation is sqrt ( 55% x 45% x 62) = 3.91 so 1.96 standard deviations is 7.82. Anything within 7.82 of the expected mean is a reasonable result. You were within 7 so yes that is a reasonable expectation.

PS: You don't have to make up the airplane story, we will help with homework anyway.