It is 0.0606
Provided the distribution is Normal, the z-score is the value such that the probability of observing a smaller value is 0.25. Thus z = -0.67449
It is 5/18.
It is 10/36 = 5/18
The answer is 0.1586
It is 84.3%
Provided the distribution is Normal, the z-score is the value such that the probability of observing a smaller value is 0.25. Thus z = -0.67449
It is 0.6915
It is 5/18.
It is 1/12.
It is 10/36 = 5/18
The answer is 0.1586
It is 84.3%
There is no real relationship. Probabilities for the Normal distribution are extremely difficult to work out. The z-score is a method used to convert any Normal distribution into the Standard Normal distribution so that its probabilities can be looked up in tables easily. There are infinitely many types of continuous probability distributions and the Normal is just one of them.
It means that your raw score is four standard deviations below the mean. This will mean different things depending on the context of the question. If you're looking at the probability of a single score occurring in a given distribution (say, a score of 40 in a distribution of scores with a mean of 80 and a std. dev. of 10), then this means that the probability of getting a 40 is very, very low--less than .00002.
The p-value is the probability of any event or the level of significance for any statistical test. The z-score is a transformation applied to a Random Variable with any Normal distribution to the Standard Normal distribution.
Pr(Z > -2) = 97.725%
There is not enough information in the question to determine if the t-distribution is the appropriate model to use. If it is, then, with, a sample size of 95 the z-score for the Gaussian distribution is a suitable approximation. The probability is 0.199, approx.