1) First, let's solve the part a of the first problem. We need to find the probability that Neil takes exactly five shots before he misses one. We know Neil's accuracy rate is 0.91. This problem can be modeled as a geometric distribution problem. In a geometric distribution, the probability that the first failure occurs on the nth trial is (p^(n-1))*(1-p), where p is the success probability. Here, p is Neil's accuracy rate, so our idea is to use the geometric distribution formula to find the probability. Replacing n with 5 and p with 0.91, we get (0.91^4) * (1 - 0.91) = ~0.0589.
So the probability that Neil takes exactly five shots before he misses one is approximately 0.0589 or 5.89%.
Now, let's move onto part b. We want to find the expected waiting time before he misses a shot. The expected value, or mean, of a geometric distribution is 1/q, where q is the failure rate.
In our case, the failure rate is 1 - p, or 1 - accuracy_rate, which is 1 - 0.91 = 0.09. So, the expected waiting time before he misses a shot is 1 / 0.09, which is approximately 11.11. So Neil is expected to take around 11 shots before he misses one.
2) For the second problem, we need to calculate what percent of the population scored better than Susan. The score of Susan is 695, the mean score is 420, and the standard deviation is 110.
To standardize Susan's score, we will calculate the z-score. The z score is (x - μ) / σ, where x is the score, μ is the mean, and σ is the standard deviation. Let's replace x with Susan's score (695), μ with the mean score (420), and σ with the standard deviation (110). This gives us a z score of (695 - 420) / 110 = 2.5.
The z-score tells us that Susan's score is 2.5 standard deviations above the mean. To find out what percent of the population scored better than Susan, we need to find the area to the right of the z score of 2.5 on the standard normal distribution. This is the same as 1 minus the cumulative area up to the z score.
So, we find the cumulative probability for a z-score of 2.5, which we would usually use a statistical table for. The result is approximately 0.9938. So the percentage of population who scored lower than Susan is 99.38%.
Therefore, the percentage of the population that scored better than Susan is 1 - 0.9938 = 0.0062 or 0.62%. So, only about 0.62% of the population scored better than Susan.